Indeterminate Forms and Newton's Method

Derivatives can help with limits and nonlinear equations, but neither operation becomes unconditional. A limit theorem requires hypotheses; an iterative method requires a starting point and convergence analysis. Getting a plausible decimal is not the same as having either guarantee.

L’Hôpital’s rule

One useful version is this: suppose $f$ and $g$ are differentiable on a deleted one-sided neighborhood of $a$, $g'$ is nonzero there, and both $f$ and $g$ tend to zero, or both have infinite magnitude. If $f'/g'$ tends to a finite or extended-real limit $L$, then $f/g$ tends to $L$. Corresponding versions apply to limits at infinity. For a two-sided result, establish the necessary behavior from both sides. The theorem follows from a generalized mean value theorem; it is a statement about a quotient of derivatives, not the derivative of the quotient.

Thus

$$\lim_{x\to0}\frac{e^x-1}{x} =\lim_{x\to0}\frac{e^x}{1}=1.$$

This is a consistency check after exponential differentiation has been established, not an independent proof of the normalization used earlier. Using a theorem to prove one of its own prerequisites would be circular.

For $\lim_{x\to\infty}\log x/x$, both numerator and denominator grow without bound and the derivative ratio is $1/x\to0$. In contrast, applying the rule to $(x^2+1)/x$ at zero is invalid: the numerator tends to $1$, not zero or infinity. The original quotient has opposite unbounded one-sided behavior, while the derivative ratio $2x$ tends to zero. The missing hypothesis changes the answer.

Rewrite other forms first

A product of a vanishing factor and an unbounded factor is not a quotient to which the rule directly applies. For $x\log x$ as $x\to0^+$, write $(\log x)/(1/x)$. The derivative ratio is $(1/x)/(-1/x^2)=-x\to0$, so the product tends to zero.

For a variable power, logarithms can expose a quotient. If $y=x^x$ for $x>0$, then $\log y=x\log x\to0$, so $y\to1$. Do not treat $0^0$ as a numerical evaluation; the particular limiting expression has been analyzed.

Newton’s tangent update

To solve $f(x)=0$, approximate the graph at a current guess $x_n$ by its tangent $f(x_n)+f'(x_n)(x-x_n)$. The tangent’s zero is

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},$$

provided $f'(x_n)\ne0$. For $f(x)=x^2-2$, the update is $(x_n+2/x_n)/2$. Starting at $x_0=1$ gives $1.5$, $1.416666\ldots$, and $1.414215686\ldots$, approaching $\sqrt2$.

Near a simple root $r$, a twice continuously differentiable function with $f'(r)\ne0$ has a neighborhood in which suitably close starting guesses converge quadratically. Roughly, the next error is proportional to the square of the present error. This is a local theorem, not a guarantee from every starting value.

For $f(x)=x^3-2x+2$, starting at zero gives $x_1=1$ and $x_2=0$, an exact two-cycle. At a point with zero derivative, the displayed update may be undefined. For functions with several roots, a different starting guess can approach a different root or fail to converge.

Certifying an approximation

A small residual $|f(x_n)|$ need not mean small distance to a root if the graph is flat. If a root $r$ and estimate $x_n$ lie in an interval where $|f'|\geq m>0$, MVT yields

$$|x_n-r|\leq|f(x_n)|/m.$$

Bisection offers a bracket-based error guarantee without requiring a nonzero derivative. A robust numerical procedure can retain a sign-changing bracket and accept Newton steps only when they stay within it, falling back to bisection otherwise. Root approximation and root existence remain separate pieces of the argument.

Exercises

1. Evaluate $\lim_{x\to0}(\sin x-x)/x^3$ by repeated justified applications of L’Hôpital’s rule.
2. Evaluate $\lim_{x\to\infty}x/e^x$.
3. Explain why the rule cannot be applied directly to $\sin x/(1+x)$ at zero.
4. Take two Newton steps for $x^2-3=0$ starting at $2$.
5. Verify the two-cycle for $x^3-2x+2$ starting at zero.
6. If $|f'|\geq2$ between an estimate and a known root and the residual is $0.006$, bound root error.

Worked solutions

1. The derivative ratios are $(\cos x-1)/(3x^2)$, $-\sin x/(6x)$, and $-\cos x/6$, yielding $-1/6$. Each earlier stage is $0/0$, with nonzero derivative denominator on each deleted side.
2. The $\infty/\infty$ form gives derivative ratio $1/e^x\to0$.
3. The denominator tends to $1$, so ordinary quotient limits give zero. The required indeterminate form is absent.
4. $x_1=2-(4-3)/4=7/4$; $x_2=(7/4+12/7)/2=97/56\approx1.732142857$.
5. $f(0)=2$, $f'(0)=-2$, so the next value is $1$. Then $f(1)=1$, $f'(1)=1$, returning zero.
6. At most $0.003$.