Limits and One-Sided Behavior

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.