A position sensor records where a cart is at successive times. It can tell us how far the cart moved between two readings. A speedometer appears to answer a more demanding question: how fast the cart is moving at one instant. Dividing distance by elapsed time seems to require two instants. Calculus will make that tension precise rather than disguise it with a sufficiently small number.

For a deliberately simple model, let the cart’s position be $s(t)=t^2$ metres, where $t$ is measured in seconds and $t\geq0$. The coefficient of $t^2$ therefore has units metres per second squared. The formula is a model, not a claim about every cart. Between $t=2$ and $t=5$, the displacement is $25-4=21$ metres and the elapsed time is $3$ seconds. The average velocity is $7$ metres per second. It does not follow that the cart moved at that velocity throughout the interval.

Functions include their domains

A real-valued function assigns exactly one real output to each allowed input. Its domain is part of its specification. The formulas $x^2$ on $\mathbb R$ and $t^2$ on $[0,\infty)$ use the same algebra but represent different functions when domains matter.

Three restrictions recur throughout this book. A denominator cannot be zero. The argument of a real square root must be nonnegative. The argument of a real logarithm must be positive. For example,

$$f(x)=\frac{\sqrt{x+2}}{x-1} \quad\hbox{has domain}\quad [-2,1)\cup(1,\infty).$$

Writing $\sqrt{x+2}$ requires $x\geq-2$; division requires $x\ne1$. Both restrictions apply. A graphing program that silently omits the forbidden point has not removed the restriction.

Composition also carries restrictions. If $g(x)=\log x$ and $f(u)=1/(u-2)$, then $(f\circ g)(x)=1/(\log x-2)$ requires $x>0$ and $x\ne e^2$. We must check the input to each operation, in the order in which it is performed.

The difference quotient

For distinct inputs $a$ and $a+h$, the average rate of change is

$$\frac{f(a+h)-f(a)}h,\qquad h\ne0.$$

The numerator is the output change and the denominator the input change. On a graph, this is the slope of the line through $(a,f(a))$ and $(a+h,f(a+h))$, called a secant line. If the output is position and the input is time, its units are distance per time. If output is cost and input is production, the units are currency per item. A slope without its variable definitions can be mathematically correct and practically useless.

For $s(t)=t^2$, expand before simplifying:

$$\frac{s(a+h)-s(a)}h =\frac{a^2+2ah+h^2-a^2}h=2a+h\qquad(h\ne0).$$

At $a=2$, intervals of length $1$, $0.1$, and $0.01$ give average velocities $5$, $4.1$, and $4.01$. Backward intervals use negative $h$: $h=-0.1$ compares times $1.9$ and $2$, giving $3.9$. Both directions suggest $4$. The original quotient remains undefined at $h=0$, however. The simplified expression describes its behavior where the quotient is defined; it does not authorize division by zero.

Direction matters

Average velocity is displacement divided by elapsed time. Average speed is total distance travelled divided by elapsed time. A person who walks three metres east and three metres west in four seconds has zero average velocity and average speed $6/4=1.5$ metres per second. The endpoint positions alone determine displacement; they do not determine total distance.

For a graph that is a straight line $f(x)=mx+b$, every secant slope is $m$. For a curved graph the interval matters. Calculus asks whether there is a well-defined local rate as the interval contracts. It does not begin by assuming that every curve has one.

Algebraic preparation

The identity $(u+v)^2=u^2+2uv+v^2$ explains the cancellation above. Factoring and conjugates provide two more tools. Since $x^2-a^2=(x-a)(x+a)$, a quotient can simplify away from $x=a$. Since $(\sqrt{x}-\sqrt a)(\sqrt{x}+\sqrt a)=x-a$, multiplying by a conjugate can expose a hidden factor. In either case, preserve the original domain while manipulating the expression.

A useful check is to calculate the quotient at one ordinary permitted input before and after simplifying. Agreement at one input is not proof of an identity, but disagreement immediately exposes an error.

Exercises

1. Determine the real domain of $\sqrt{9-x^2}/(x-2)$.
2. For $p(t)=3t^2-2t+1$ metres, calculate average velocity from $t=1$ to $t=3$ seconds.
3. Simplify $[p(a+h)-p(a)]/h$ and state its restriction.
4. A tank contains $V(t)=100-4t$ litres for $0\leq t\leq25$ minutes. Interpret its average rate over any nonzero interval in this domain.
5. Explain why cancelling $x-1$ in $(x^2-1)/(x-1)$ does not define the original function at $1$.
6. A car returns to its starting position after an hour. What can be concluded about its average velocity and average speed?

Worked solutions

1. The square root requires $-3\leq x\leq3$ and the denominator excludes $2$, giving $[-3,2)\cup(2,3]$.
2. $p(1)=2$ and $p(3)=22$, so the average is $(22-2)/(3-1)=10$ metres per second.
3. Expansion gives $6a+3h-2$, valid for $h\ne0$ and permitted times. The terms $3a^2-2a+1$ cancel, leaving $h(6a+3h-2)$.
4. The rate is $-4$ litres per minute. The negative sign indicates loss of volume. The model stops when the tank empties; extrapolating to negative volume is outside its domain.
5. The two expressions agree for $x\ne1$. The expression $x+1$ has a value at $1$ that the original quotient lacks. Extending a function is a separate definition.
6. Average velocity is zero. Average speed is nonnegative and positive if any nonzero distance was travelled. Its value requires the total distance.

The secant calculation gave a quantity that was undefined at the input we wanted. This is not an unusual defect to patch. It is the reason we need a new concept. We want to describe what outputs do when inputs approach a point through other allowed inputs.

Consider

$$f(x)=\frac{x^2-4}{x-2},\qquad x\ne2.$$

Factoring shows that $f(x)=x+2$ at every input in its domain. As $x$ approaches $2$, the outputs approach $4$. We write $\lim_{x\to2}f(x)=4$. We have made a statement about values near $2$, although $f(2)$ is not defined.

A punctured neighborhood

The informal meaning of a finite limit $L$ is that outputs can be made arbitrarily close to $L$ by keeping inputs sufficiently close to $a$, while excluding $a$ itself. The exclusion matters. A punctured neighborhood has the form $0<|x-a|<\delta$. For the two-sided limits used here, our functions are defined throughout some such neighborhood. More general domain-relative limits are possible, but we will state explicitly when an endpoint requires a one-sided approach.

Now define a new function $g$ that agrees with $f$ for $x\ne2$ but has $g(2)=100$. Its limit at $2$ is still $4$. Altering one value does not change nearby behavior. On the other hand, altering all values to the right of $2$ can change the limit even if the value at $2$ stays fixed.

The assertion $\lim_{x\to a}f(x)=L$ also does not require $f(x)$ to equal $L$ near $a$. For $f(x)=x$, the limit at $0$ is $0$, but no nonzero input has output $0$. Nor does it forbid equality: a constant function can equal its limit everywhere.

Left and right must agree

The notation $x\to a^-$ restricts the approach to $x<a$; $x\to a^+$ restricts it to $x>a$. A finite two-sided limit exists exactly when the two finite one-sided limits exist and agree, under the neighborhood assumption above.

For a concrete jump, let $q(x)=x+1$ when $x<0$ and $q(x)=x+3$ when $x\geq0$. Then

$$\lim_{x\to0^-}q(x)=1,\qquad \lim_{x\to0^+}q(x)=3.$$

No two-sided finite limit exists at $0$. Choosing $q(0)=1$, $2$, or $3$ cannot repair this disagreement. There is no single output target that describes both sides.

At a domain endpoint, a one-sided question is often the appropriate one. For $\sqrt{x}$ on $[0,\infty)$, the right-hand limit at $0$ is $0$. There is no real-valued graph to its left. We should state the right-hand limit rather than pretend we have checked both sides.

A table is evidence, not a guarantee

For $f(x)=(x^2-4)/(x-2)$, inputs $1.9$, $1.99$, $2.01$, and $2.1$ give $3.9$, $3.99$, $4.01$, and $4.1$. The factorization establishes what every permitted nearby input does; the table only illustrates a few inputs. A finite collection of samples cannot rule out a narrow spike or rapid oscillation between sampled points.

For example, $\sin(1/x)$ has no limit as $x\to0$. Along $x_n=1/(\pi/2+2\pi n)$ its values are $1$. Along $y_n=1/(3\pi/2+2\pi n)$ they are $-1$. Both input sequences tend to zero. A common finite limit would force both output sequences toward the same value, which is impossible.

Contrast this with $x\sin(1/x)$. Its oscillations are trapped in an envelope whose height tends to zero. That function does have limit zero. Oscillation by itself does not settle the question; its amplitude matters.

Infinity describes behavior

Writing $\lim_{x\to0}1/x^2=+\infty$ means the values eventually exceed any prescribed positive bound as nonzero inputs approach zero. It does not assign a real value called infinity. For $1/x$, the right-hand behavior is $+\infty$ and the left-hand behavior is $-\infty$, so even their unbounded directions disagree.

Limits at infinity concern large inputs instead: $\lim_{x\to\infty}1/x=0$. The graph approaches the horizontal axis while never reaching it. A vertical asymptote concerns behavior near a finite input; a horizontal asymptote concerns behavior as inputs become unbounded. They answer different questions.

Exercises

1. Let $f(x)=x+5$ for $x\ne1$ and $f(1)=-8$. Find the limit at $1$ and the function value there.
2. Find both one-sided limits of $|x|/x$ at zero. Does a two-sided limit exist?
3. Explain why $1/x^2$ has no finite real limit at zero.
4. Give two functions that have the same limit at $3$ but different values at $3$.
5. Explain why a graph drawn from 1,000 sample points cannot establish a limit for every function.
6. Find two sequences showing that $\cos(1/x)$ has no limit at zero.

Worked solutions

1. The limit is $6$ because the nearby formula is $x+5$; $f(1)=-8$.
2. The left limit is $-1$ and the right limit is $1$, so no two-sided limit exists.
3. Values exceed every real bound near zero. No finite candidate can remain close to those arbitrarily large outputs.
4. Use $f(x)=x$ everywhere and $g(x)=x$ except $g(3)=7$. Both limits are $3$.
5. Sampling leaves unexamined inputs. A function can agree at all sampled points but behave differently between them or closer to the target.
6. For positive integers $n$, use $x_n=1/(2\pi n)$ and $y_n=1/((2n+1)\pi)$. Outputs are $1$ and $-1$ while both sequences approach zero.

Substitution is often the last line of a limit calculation, but it is not the definition of a limit. It works for polynomials because their limits obey algebraic rules; it works for a rational expression at points where the denominator has a nonzero limit. Recognizing those conditions lets us distinguish an ordinary calculation from one that needs more work.

The algebra of finite limits

Suppose $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$, with $L,M$ finite. Then limits commute with sums, differences, constant multiples, and products. Thus $\lim(f+g)=L+M$ and $\lim(fg)=LM$. For quotients we additionally require $M\ne0$, giving $\lim(f/g)=L/M$. Positive integer powers follow by repeated products.

For composition, if $f(x)\to L$ and $F$ is continuous at $L$, then $F(f(x))\to F(L)$, provided the composition is defined nearby. Continuity will receive a careful definition shortly. Root functions are continuous on their real domains, with one-sided care at endpoints. Hence

$$\lim_{x\to2}\frac{3x^2-x+1}{x+4}=\frac{11}{6}.$$

The denominator approaches $6$, not zero. That single check justifies the final division.

Zero over zero is a prompt to investigate

For $(x^2-9)/(x-3)$ at $3$, substitution gives the meaningless quotient $0/0$. It does not give an answer of zero, infinity, or “undefined limit.” For $x\ne3$,

$$\frac{x^2-9}{x-3}=x+3,$$

so the limit is $6$. The two expressions agree throughout a punctured neighborhood. Limits depend on exactly that nearby behavior.

Compare $x/x$, $x^2/x$, and $x/x^2$ at zero. All display $0/0$ on substitution, yet their behaviors are respectively a constant $1$, a limit $0$, and incompatible unbounded one-sided limits. The notation indeterminate form records insufficient information, not a result.

Conjugates handle some square-root cases. For $x\ne4$ near $4$,

$$\frac{\sqrt{x}-2}{x-4} =\frac1{\sqrt{x}+2},$$

because multiplying numerator and denominator by $\sqrt{x}+2$ exposes $x-4$. The limit is $1/4$. Multiplying by the conjugate is legal nearby because $\sqrt{x}+2$ is nonzero.

Squeezing every nearby value

The squeeze theorem says that if $u(x)\leq f(x)\leq v(x)$ for all sufficiently near allowed inputs, and both bounding functions approach $L$, then $f(x)$ also approaches $L$. The inequalities must hold throughout the relevant neighborhood, not only at selected table entries.

Since $|\sin(1/x)|\leq1$, we have

$$-|x|\leq x\sin(1/x)\leq |x|.$$

Both outer expressions tend to zero. This supplies a conclusion even though the middle function oscillates infinitely often near zero.

An important trigonometric limit is $\lim_{x\to0}\sin x/x=1$, with angles measured in radians. On $0<x<\pi/2$, the unit-circle area comparison gives $\sin x<x<\tan x$. Dividing yields $\cos x<\sin x/x<1$. As $x\to0^+$, both bounds tend to $1$; symmetry gives the left side. This argument uses elementary geometric continuity of cosine, not a derivative formula for sine. Also,

$$\frac{1-\cos x}{x}=\frac{\sin x}{x}\frac{\sin x}{1+\cos x}\longrightarrow0.$$

These results will justify trigonometric derivatives. Degrees would introduce a conversion factor; the radian convention is mathematical infrastructure, not a cosmetic preference.

Unbounded inputs and dominant terms

To study a rational expression as $x\to\infty$, divide numerator and denominator by a suitable highest power. For example,

$$\frac{5x^2+2x-1}{2x^2+7} =\frac{5+2/x-1/x^2}{2+7/x^2}\longrightarrow\frac52.$$

If the denominator degree is larger, the ratio tends to zero. If the numerator degree is larger, polynomial growth remains; its sign and direction must still be checked. A memorized degree rule should summarize this calculation rather than replace it.

For radicals, $\sqrt{x^2}=|x|$, not always $x$. Therefore $\sqrt{x^2+1}/x$ tends to $1$ as $x\to+\infty$ and $-1$ as $x\to-\infty$. Forgetting the absolute value changes the second answer.

Exercises

1. Evaluate $\lim_{x\to2}(x^2+x-6)/(x-2)$.
2. Evaluate $\lim_{x\to0}(\sqrt{1+x}-1)/x$.
3. Find $\lim_{x\to0}\sin(5x)/x$ and justify the factor $5$.
4. Use squeezing to evaluate $\lim_{x\to0}x^2\cos(1/x)$.
5. Find the limits of $(3x^3-x)/(x^3+2)$ at positive and negative infinity.
6. Explain why the quotient law cannot establish $\lim_{x\to0}\sin x/x$ directly.
7. Find $\lim_{x\to-\infty}\sqrt{4x^2+1}/x$.
8. Give an example showing that $\lim(fg)$ can exist even when neither factor has a finite limit.

Worked solutions

1. Factor the numerator as $(x+3)(x-2)$; the limit is $5$.
2. The conjugate gives $1/(\sqrt{1+x}+1)$ near zero, with limit $1/2$.
3. Write $5[\sin(5x)/(5x)]$. The bracket tends to $1$ in radians, giving $5$.
4. The absolute value is bounded by $x^2$, so the limit is zero.
5. Dividing by $x^3$ gives $3$ in either direction. The nonzero denominator limit justifies the quotient step.
6. The denominator limit is zero, violating that law’s nonzero-denominator hypothesis.
7. Write $|x|\sqrt{4+1/x^2}/x$. For negative $x$, $|x|/x=-1$, giving $-2$.
8. Let $f(x)=1/x$ and $g(x)=x$ for $x\ne0$. Their product has limit $1$ at zero, although the first factor has no finite limit there. To make neither factor converge, take $f(x)=2+\sin(1/x)$ and $g(x)=1/f(x)$; their product is identically $1$ while both factors oscillate.

A numerical table can make a limit plausible. A proof must handle every input sufficiently near the target, including inputs that no table contains. We therefore need a statement that says exactly what “sufficiently near” must accomplish.

Let $f$ be defined throughout a punctured neighborhood of $a$. We say $\lim_{x\to a}f(x)=L$ if

$$\text{for every }\varepsilon>0\text{ there exists }\delta>0 \text{ such that }0<|x-a|<\delta \implies |f(x)-L|<\varepsilon.$$

Here $\varepsilon$ is the requested output tolerance and $\delta$ is the promised input tolerance. We receive an arbitrary positive $\varepsilon$, then choose a suitable $\delta$, then guarantee the implication for every permitted $x$. A choice of $\delta$ may depend on $\varepsilon$, $a$, and the fixed function. It must not depend on an input $x$ selected afterward.

A linear proof from start to finish

Prove $\lim_{x\to2}(3x-1)=5$. The output error is

$$|(3x-1)-5|=3|x-2|.$$

For a given $\varepsilon>0$, choose $\delta=\varepsilon/3$. If $0<|x-2|<\delta$, then $|(3x-1)-5|=3|x-2|<3\delta=\varepsilon$. Since this works for every positive tolerance, the proof is complete.

Notice the distinction between scratch work and the proof. We discovered $\delta$ by working backward from the desired inequality. We presented the proof forward, starting with an arbitrary tolerance and the explicit choice. A successful proof does not require the largest possible $\delta$. Any positive choice that ensures the implication works.

A quadratic needs a local bound

To prove $\lim_{x\to2}x^2=4$, factor the error:

$$|x^2-4|=|x-2||x+2|.$$

The second factor depends on $x$. We first arrange $|x-2|<1$, which gives $1<x<3$ and hence $|x+2|<5$. Now choose

$$\delta=\min\{1,\varepsilon/5\}.$$

If $0<|x-2|<\delta$, then $|x^2-4|<5|x-2|<5\delta\leq\varepsilon$. The final strict inequality comes from the strict bound on $|x-2|$. The minimum enforces two requirements simultaneously: remain in a neighborhood where the auxiliary factor is controlled, and make the final error small.

For a general fixed real $a$, $|x-a|<1$ implies $|x+a|<2|a|+1$. Thus $\delta=\min\{1,\varepsilon/(2|a|+1)\}$ proves $\lim_{x\to a}x^2=a^2$.

The largest symmetric tolerance for a square

The laboratory uses an exact bound to make the geometry visible. Put $x=a+h$. For $|h|<\delta$,

$$|x^2-a^2|=|2ah+h^2| \leq2|a||h|+h^2 <2|a|\delta+\delta^2.$$

The supremum of the output error on this open neighborhood is $2|a|\delta+\delta^2$, approached from the outward direction. A sufficient and, for the symmetric neighborhood, maximal choice is

$$\delta_* = \sqrt{a^2+\varepsilon}-|a| =\frac{\varepsilon}{\sqrt{a^2+\varepsilon}+|a|}.$$

The second form avoids subtracting nearly equal numbers in computation. At $a=0$, this becomes $\sqrt\varepsilon$. At $a=2$ and $\varepsilon=0.1$, it is approximately $0.0248457$. Setting $\delta=\delta_*$ is valid: the boundary supremum equals $\varepsilon$, but boundary inputs are excluded and every interior error is strictly smaller. This is a useful example of why strict and nonstrict inequalities cannot be interchanged casually.

How to show a candidate fails

Negating the definition gives: there exists some $\varepsilon_0>0$ such that for every $\delta>0$, there is a permitted $x$ with $0<|x-a|<\delta$ and $|f(x)-L|\geq\varepsilon_0$.

For $f(x)=|x|/x$, the candidate $L=1$ fails at zero. Choose $\varepsilon_0=1$. Whatever $\delta>0$ is offered, choose $x=-\delta/2$. Then $f(x)=-1$ and $|f(x)-1|=2\geq1$. To rule out every candidate, the one-sided disagreement established previously is more economical: any two-sided limit would equal both $-1$ and $1$.

A finite limit is unique

Suppose two distinct finite limits $L$ and $M$ existed. Let $\varepsilon=|L-M|/3$. There would be neighborhoods in which outputs are within $\varepsilon$ of each limit. Choose a noncentral input inside both neighborhoods. The triangle inequality would give

$$|L-M|\leq|L-f(x)|+|f(x)-M|<2|L-M|/3,$$

a contradiction. The existence of nearby domain points matters; our punctured-neighborhood assumption supplies them. This proof establishes uniqueness without relying on a picture.

Exercises

1. Prove $\lim_{x\to-1}(5x+2)=-3$ with an explicit $\delta$.
2. Prove $\lim_{x\to0}x^2=0$ using the largest symmetric tolerance.
3. Prove $\lim_{x\to3}x^2=9$ using a local bound of radius one.
4. Explain why choosing $\delta=\varepsilon/|x+2|$ is not an acceptable construction in the quadratic proof.
5. For $a=1$, $\varepsilon=0.25$, does $\delta=0.2$ guarantee the square-function tolerance? Find the supremum error.
6. Show that the candidate limit $L=0$ fails for the constant function $f(x)=2$ at zero.

Worked solutions

1. Choose $\delta=\varepsilon/5$. The output error is $5|x+1|<\varepsilon$.
2. Choose $\delta=\sqrt\varepsilon$. Then $0<|x|<\delta$ implies $x^2<\varepsilon$.
3. If $|x-3|<1$, then $|x+3|<7$. Choose $\min\{1,\varepsilon/7\}$ and factor $|x^2-9|$.
4. This choice depends on the later input $x$. We must provide a single radius that works for every input inside it; bounding $|x+2|$ by a fixed constant accomplishes that.
5. The supremum is $2(1)(0.2)+0.2^2=0.44$, greater than $0.25$, so the guarantee fails. For example, $x=1.19$ is allowed and produces error $0.4161$.
6. Choose $\varepsilon_0=1$. Every nonzero input within every proposed positive radius has error $2\geq1$.

A thermostat reading can pass from below a target to above it. If temperature varies continuously, it must equal the target somewhere between the readings. This conclusion concerns an unobserved instant. It follows from a mathematical assumption about the entire interval, not from the two readings alone. A sensor that jumps between quantized readings need not display every intermediate number.

Matching nearby behavior to the actual value

A function is continuous at $a$ when $f(a)$ exists, the finite limit $\lim_{x\to a}f(x)$ exists, and that limit equals $f(a)$. For an interior point these are two-sided conditions. At an included endpoint of an interval, use the appropriate one-sided limit. Continuity on $[a,b]$ means continuity at all interior points, right continuity at $a$, and left continuity at $b$.

For $f(x)=(x^2-1)/(x-1)$, the limit at $1$ is $2$ but the function is undefined there. Defining $f(1)=2$ creates a continuous extension. Defining $f(1)=7$ leaves a removable discontinuity: changing one value can repair it. A jump with different left and right limits cannot be repaired that way. Neither can the unbounded behavior of $1/x$ at zero.

Polynomials are continuous everywhere on the real line. Rational functions are continuous wherever their denominators are nonzero. Compositions of continuous functions are continuous where the composition is defined. These are consequences of limit laws. They let us verify continuity without repeating a tolerance proof at every point.

Joining two formulas

Let $f(x)=kx+1$ for $x<2$ and $f(x)=x^2-1$ for $x\geq2$. Each piece is continuous on its own interval. At the join, the left limit is $2k+1$, the right limit is $3$, and $f(2)=3$. Continuity requires $2k+1=3$, hence $k=1$. Checking only that both formulas are polynomials misses the join, which is the only point that needs extra work.

The intermediate value theorem

If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then some $c\in[a,b]$ satisfies $f(c)=N$. If $N$ lies strictly between the endpoint values, such a $c$ lies in $(a,b)$. The theorem guarantees existence. It does not guarantee one solution, a formula for it, or that the function is monotone.

For $p(x)=x^3+x-1$, continuity holds on $[0,1]$, and $p(0)=-1$, $p(1)=1$. Therefore a zero lies in $(0,1)$. It is unique because for $v>u$,

$$p(v)-p(u)=(v-u)(v^2+uv+u^2+1)>0.$$

The second factor is positive since $v^2+uv+u^2=(v+u/2)^2+3u^2/4\geq0$. We supplied a separate monotonicity argument; IVT alone would not establish uniqueness.

The continuity hypothesis is essential. A function equal to $-1$ for $x<0$ and $1$ for $x\geq0$ changes sign across $[-1,1]$ without ever equalling zero. A sign change is not a substitute for continuity.

Bisection with an error bound

If a continuous function has opposite signs at the endpoints, evaluate its midpoint. Keep the half whose endpoints still bracket a sign change, unless the midpoint is already a root. Repeating gives nested intervals. After $n$ halvings, the bracket length is $(b-a)/2^n$, and its midpoint is within $(b-a)/2^{n+1}$ of at least one root remaining in the bracket.

For $p$ above, $p(0.5)=-0.375$, while $p(0.75)=0.171875$. A root therefore lies in $(0.5,0.75)$. The midpoint $0.625$ has a guaranteed error at most $0.125$ at this stage. To guarantee midpoint error no greater than $0.001$ from the initial unit interval, nine halvings suffice because $1/2^{10}<0.001$. A small function residual alone is a different notion of error; a very flat function can have a small residual far from a root.

The extreme value theorem

If $f$ is continuous on a closed bounded interval $[a,b]$, it attains an absolute maximum and an absolute minimum on that interval. The conclusion is stronger than having a finite upper or lower bound: actual inputs achieve the extreme outputs.

Both interval conditions matter. On $(0,1)$, the continuous function $f(x)=x$ attains neither its supremum $1$ nor its infimum $0$. On $\mathbb R$, the continuous function $x^2$ has no absolute maximum. On $[0,1]$, a discontinuous function can be unbounded: set $f(0)=0$ and $f(x)=1/x$ for $x>0$.

IVT and EVT are foundational real-number theorems. Their full proofs depend on completeness of the real numbers. Here we use their stated conclusions and identify the hypotheses each application needs. Later, EVT will ensure optimization problems have candidates worth seeking.

Exercises

1. Find $k$ making $f(x)=kx+2$ for $x<1$, $f(x)=x^2+4$ for $x\geq1$ continuous.
2. Prove that $x^3-2$ has a zero in $(1,2)$. Is uniqueness supplied by IVT?
3. How many bisection halvings reduce an initial bracket of length $4$ to midpoint error at most $0.01$?
4. Give a continuous function on an open bounded interval with no maximum.
5. Does EVT imply that a maximum occurs at an interior point?
6. For $f(x)=1/x$ on $[-1,1]\setminus\{0\}$, explain why opposite endpoint signs do not establish a zero.

Worked solutions

1. The right value and limit at $1$ are $5$; $k+2=5$ gives $k=3$.
2. The polynomial is continuous and has endpoint values $-1$ and $6$. IVT establishes a root; uniqueness follows separately from strict increase of the cube function.
3. Require $4/2^{n+1}\leq0.01$. Eight halvings give $4/512=0.0078125$; seven give $0.015625$, so eight suffice.
4. $f(x)=x$ on $(0,1)$ has supremum $1$ that is never attained.
5. No. For $f(x)=x$ on $[0,1]$, the maximum is at $1$.
6. The function is not continuous on the full closed interval; it is undefined at zero. IVT does not apply.

For the cart with position $s(t)=t^2$, the average velocity from $a$ to $a+h$ was $2a+h$. Both positive and negative increments approach the same value $2a$. We can now define the instantaneous velocity without assigning a meaning to division by zero: take the limit of the nonzero-interval average velocities.

For a function defined on an open interval containing $a$, the derivative at $a$ is the finite limit

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

if it exists. An equivalent form uses $x=a+h$ and lets $x\to a$. Both formulations compare the output at the target with outputs at nearby inputs. We use a two-sided limit for interior differentiability. Endpoint derivatives, when useful, will be named one-sided derivatives.

A quadratic from the definition

Let $f(x)=2x^2-3x+1$. Expanding gives

$$\frac{f(a+h)-f(a)}h =\frac{4ah+2h^2-3h}h=4a+2h-3,\qquad h\ne0.$$

Therefore $f'(a)=4a-3$. At $a=2$, the function value is $3$ and derivative is $5$. The tangent line has equation

$$y=f(2)+f'(2)(x-2)=3+5(x-2).$$

A tangent is specified by its local limiting slope. It need not stay on one side of the curve or touch it at only one point. The tangent $y=0$ to $x^3$ at zero crosses the curve there.

Derivatives carry units

If $f$ measures temperature in degrees Celsius and $x$ measures hours, $f'(a)$ has units degrees Celsius per hour. A derivative of $-2$ means temperature is locally decreasing at that rate; it does not mean the temperature is negative. If position is measured in metres and time in seconds, velocity has units metres per second and acceleration, the derivative of velocity, has units metres per second squared.

For $s(t)=t^3-3t$ on a specified time interval, $v(t)=3t^2-3$. Positive velocity indicates motion in the positive coordinate direction. Speed is $|v(t)|$, so negative velocity is not negative speed. Later we will integrate speed to compute total distance.

Continuous does not mean differentiable

For $f(x)=|x|$ at zero, the difference quotient is $|h|/h$. It equals $-1$ for negative $h$ and $1$ for positive $h$. Thus the two-sided derivative does not exist even though $|x|\to0=f(0)$. The graph has a corner.

The function $x^{1/3}$ at zero has quotient $h^{-2/3}$, which grows without bound. This is often described geometrically as a vertical tangent. It is not a finite derivative under our definition. A discontinuity also prevents differentiability, but a smooth-looking numerical plot is not proof that discontinuities or corners are absent.

Why differentiability forces continuity

Suppose $f'(a)$ exists and is finite. For $h\ne0$,

$$f(a+h)-f(a)=h\frac{f(a+h)-f(a)}h.$$

As $h\to0$, the first factor tends to zero and the second tends to $f'(a)$. The product tends to zero. Hence $f(a+h)\to f(a)$, establishing continuity at $a$. Finiteness of the derivative matters because the ordinary product limit law needs finite limits.

This proof gives a useful diagnostic: if continuity fails, a finite derivative cannot exist. Its converse remains false, as the absolute-value example shows. A theorem and its converse are separate claims.

The local linear model

Rearrange the defining quotient into

$$f(a+h)=f(a)+f'(a)h+r(h),\qquad \frac{r(h)}h\longrightarrow0.$$

The derivative supplies a linear change, and the residual is small relative to the increment, not merely small in absolute size. For $f(x)=x^2$, $r(h)=h^2$, so $r(h)/h=h\to0$. This relative-error statement distinguishes a tangent approximation from any arbitrary line passing through the point.

The secant laboratory displays exact formulas for the square, cube, and absolute-value functions. A finite list of small increments cannot prove a derivative. The associated algebra explains the limit; the table and graph help us inspect its consequences. For polynomials the program evaluates algebraically simplified quotients to avoid catastrophic subtraction at small increments.

Exercises

1. Use the definition to differentiate $f(x)=3x^2+2x$ at a general $a$.
2. Find the tangent to $x^2$ at $x=-1$.
3. Use first principles to find the derivative of $1/x$ at $a\ne0$.
4. Show that the tangent to $x^3$ at zero crosses its graph.
5. A temperature derivative is zero at noon. Must the temperature be zero? Must it remain constant afterward?
6. Explain the different reasons that $|x|$ and $x^{1/3}$ lack a finite derivative at zero.

Worked solutions

1. Expansion gives $[f(a+h)-f(a)]/h=6a+3h+2$ for nonzero $h$, hence $f'(a)=6a+2$.
2. The point is $(-1,1)$ and slope $-2$, so $y=1-2(x+1)=-2x-1$.
3. The quotient simplifies to $-1/[a(a+h)]$, provided $h\ne0$ and $a+h\ne0$. Its limit is $-1/a^2$.
4. The derivative at zero is $\lim h^2=0$, so the tangent is the horizontal axis. Cubes have opposite signs on opposite sides of zero.
5. Neither. The derivative describes an instantaneous rate, not the level or a future interval. For example, $T(t)=20+t^2$ has derivative zero at $t=0$ without being constant.
6. Absolute value has two distinct finite one-sided slopes; the cube root has a quotient unbounded in the positive direction. Neither gives the required finite two-sided limit.

A derivative at one point is a number. A derivative function assigns such numbers to all inputs where the required finite limit exists. For $f(x)=x^2$, the calculation $f'(a)=2a$ holds for every real $a$, so the derivative function is $f'(x)=2x$. The variable has changed its role from a fixed base point in one proof to the argument of a new function.

The notation $dy/dx$ also denotes the derivative when $y=f(x)$. At this stage it is notation for a limit-defined rate, not permission to cancel arbitrary symbols. Later, the chain rule and substitution will explain why certain manipulations that resemble fraction arithmetic work.

Cubes and integer powers

For $f(x)=x^3$, the difference quotient is

$$\frac{(x+h)^3-x^3}h=3x^2+3xh+h^2,\qquad h\ne0.$$

Letting $h\to0$ gives $f'(x)=3x^2$. Both positive and negative nonzero increments yield the same expression’s limit.

For a positive integer $n$, the binomial theorem gives

$$(x+h)^n=x^n+nx^{n-1}h+\sum_{k=2}^n {n\choose k}x^{n-k}h^k.$$

Subtract $x^n$ and divide by nonzero $h$. Every term of the sum contains a positive power of $h$ and tends to zero. The result is the power rule $d(x^n)/dx=nx^{n-1}$ for positive integers. The case $n=1$ is simply a constant quotient $1$. A constant function has derivative zero because its output difference is always zero. Extending the power rule to all real exponents needs additional results; the binomial argument alone does not supply that extension.

Reciprocal and square root

The reciprocal calculation gives

$$\left(\frac1x\right)'=-\frac1{x^2},\qquad x\ne0.$$

The derivative cannot include zero because the original function has no value there. For $f(x)=\sqrt{x}$ at $x>0$, conjugates give

$$\frac{\sqrt{x+h}-\sqrt{x}}h =\frac1{\sqrt{x+h}+\sqrt{x}} \longrightarrow\frac1{2\sqrt{x}}.$$

The function’s domain is $[0,\infty)$, while its ordinary two-sided derivative has domain $(0,\infty)$. Even the right-hand difference quotient at zero grows without bound, so it does not supply a finite endpoint derivative. A formula and its domain must travel together.

For $|x|$, the derivative is $-1$ on $(-\infty,0)$ and $1$ on $(0,\infty)$, and is undefined at zero. The derivative function records a change of slope and the missing derivative at the corner. Drawing a line between the two branches would add information the function does not contain.

Higher derivatives and interpretation

If $f'$ is differentiable, its derivative is $f''$, the second derivative. For $f(x)=x^3$, $f'=3x^2$, $f''=6x$, $f'''=6$, and every derivative of order four or greater is zero. The $n$th derivative is written $f^{(n)}$, while $f^n$ normally means a power; these symbols are not interchangeable.

For position $s(t)$, the first derivative is velocity and the second acceleration. Positive acceleration means velocity is increasing, but speed need not increase. If a vehicle is moving in the negative direction with $v=-10$ and $a=2$, its velocity moves toward zero and its speed initially decreases. The signs of velocity and acceleration together determine this behavior.

Reading a derivative graph

Where $f'>0$, tangent slopes are positive; where $f'<0$, they are negative. A theorem in a later chapter will justify conclusions about increase on an interval. A derivative equal to zero at one point does not itself identify a maximum or minimum: $x^3$ has derivative zero at zero and continues increasing through it.

Two different functions can have the same derivative. For instance $x^2$ and $x^2+7$ both have derivative $2x$. Differentiation discards a vertical offset. The mean value theorem will show that, on an interval, functions with the same derivative can differ only by a constant. This observation will become the foundation of antiderivatives.

Exercises

1. Differentiate $x^5$ directly by expansion, identifying the term that survives the limit.
2. Give the domains of $\sqrt{x-2}$ and its derivative.
3. Find the first three derivatives of $x^4$.
4. Describe the derivative of $|x-3|$ at every real input where it exists.
5. Give two functions with derivative $4x^3$.
6. Explain why negative velocity with negative acceleration produces increasing speed locally when both signs persist.

Worked solutions

1. The quotient is $5x^4+10x^3h+10x^2h^2+5xh^3+h^4$. Only $5x^4$ survives.
2. The function is defined on $[2,\infty)$; the finite ordinary derivative $1/(2\sqrt{x-2})$ is defined on $(2,\infty)$.
3. $4x^3$, $12x^2$, and $24x$.
4. It equals $-1$ for $x<3$, $1$ for $x>3$, and is undefined at $3$ because the one-sided slopes disagree.
5. $x^4$ and $x^4+10$ are examples.
6. Velocity becomes more negative; its magnitude therefore increases. Speed is $-v$ while $v<0$, so its derivative is $-a>0$.

First principles establish what a derivative means. They are less convenient for differentiating a complicated expression repeatedly. We can prove rules once, then use them whenever their hypotheses hold. Every rule in this chapter assumes the component functions are differentiable at the point in question. The quotient rule also requires a nonzero denominator.

Linearity comes from the limit laws

For constants $c,d$, expand the difference quotient of $cf+dg$ and split it into two quotients. Taking limits gives

$$(cf+dg)'=cf'+dg'.$$

Thus $(4x^3-2x^2+7)'=12x^2-4x$. The constant contributes zero. Differentiating the square of a function is not covered by linearity; squaring is multiplication, not addition.

Why both factors contribute

For $F=fg$, write the numerator as

$$f(x+h)g(x+h)-f(x)g(x) =f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)].$$

After division by $h$, take limits. Differentiability implies $f(x+h)\to f(x)$, so

$$(fg)'=fg'+gf'.$$

The extra term that a naive “differentiate both factors and multiply” approach loses records the change in one factor while the other remains present. For $x\cdot x$, the rule gives $x+x=2x$, whereas multiplying the two derivatives gives $1$, which fails even this basic check.

For $F(x)=(x^2+1)(x^3-2)$,

$$F'=2x(x^3-2)+(x^2+1)3x^2=5x^4+3x^2-4x.$$

Expanding first gives $F=x^5+x^3-2x^2-2$ and the same derivative. Independent representations are useful error checks. They are often more informative than repeating the same calculation with slightly different notation.

The quotient rule

For a differentiable nonzero $g$, the difference quotient for $1/g$ simplifies to

$$-\frac{g(x+h)-g(x)}{h\,g(x+h)g(x)}.$$

Its limit is $-g'/g^2$. Continuity and $g(x)\ne0$ ensure $g(x+h)$ stays nonzero for sufficiently small increments. Applying the product rule to $f(1/g)$ gives

$$\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}.$$

The order in the numerator matters. For $Q(x)=(x^2+1)/(x-1)$,

$$Q'(x)=\frac{2x(x-1)-(x^2+1)}{(x-1)^2} =\frac{x^2-2x-1}{(x-1)^2},\quad x\ne1.$$

Polynomial division gives $Q=x+1+2/(x-1)$, whose derivative is $1-2/(x-1)^2$, algebraically identical to the result. The forbidden input $1$ remains forbidden in both versions.

Negative integer powers

For a positive integer $n$, apply the reciprocal rule to $x^n$:

$$\frac d{dx}x^{-n}=-\frac{nx^{n-1}}{x^{2n}}=-nx^{-n-1},\qquad x\ne0.$$

The power rule therefore holds for all integer exponents on their relevant domains. At exponent zero the function is constant wherever it is defined. This extension came from a new argument, not from silently expanding the earlier binomial proof beyond its assumptions.

Choosing a method

First identify the outermost operation. An expression can contain sums inside a product inside a quotient. Start with the outside, retain its component functions intact, then differentiate those components. Simplify when simplification reduces error; leaving a correctly factored derivative is often preferable to an unnecessarily long polynomial.

Numerical difference quotients at ordinary inputs can detect a sign error. Very small increments may instead amplify floating-point cancellation. A numerical agreement supports a check at that input; the proved rules establish the symbolic result across the domain.

Exercises

1. Differentiate $(2x-1)(x^2+3)$ by the product rule and by expansion.
2. Differentiate $(x+1)/(x+2)$ and state its domain.
3. Differentiate $x^{-3}+5x^2-8$.
4. Explain why $(f/g)'=f'/g'$ cannot be a general rule.
5. Derive the rule for a product of three differentiable functions using the two-factor rule twice.
6. Find the tangent slope of $(x^2+1)/(x-1)$ at $x=0$.

Worked solutions

1. $2(x^2+3)+(2x-1)2x=6x^2-2x+6$. Expanding first gives $2x^3-x^2+6x-3$ and the same derivative.
2. $[(x+2)-(x+1)]/(x+2)^2=1/(x+2)^2$, for $x\ne-2$.
3. $-3x^{-4}+10x$, for $x\ne0$.
4. Take $f=g=x$ away from zero. The quotient is constant with derivative zero, while $f'/g'=1$.
5. $(fgh)'=f'gh+fg'h+fgh'$.
6. Substitute into the verified derivative: $(-1)/1=-1$.