Indeterminate Forms and L’Hôpital’s Rule

A conditional shortcut

When a quotient approaches $0/0$ or an infinite-over-infinite form, comparing derivatives can reveal relative rates of change. Under the hypotheses of L’Hôpital's theorem, if the derivative ratio has a finite or infinite limit, the original quotient has that limit too.

In a suitable one-sided punctured interval, require differentiability of numerator and denominator and a nonzero denominator derivative. Check the original indeterminate form, then evaluate the derivative ratio. The same logic extends to infinite endpoints. If the derivative ratio has no limit, the theorem gives no conclusion about the original quotient.

Visual guide

VISUAL GUIDEA removable ratio near zero

(eˣ − 1)/x, left
(eˣ − 1)/x, right
Limit y = 1

The graph of (eˣ − 1)/x approaches 1 even though substituting zero produces 0/0. L’Hôpital’s rule evaluates the derivative ratio eˣ/1 near zero; it does not say the two functions are equal away from the limit.

Worked example: two valid applications

Consider $\lim_{x\to0}(e^x-1-x)/x^2$ . Substitution gives $0/0$ . Differentiating numerator and denominator separately gives $(e^x-1)/(2x)$ , still $0/0$ . A second valid application gives $e^x/2\to1/2$ .

The operation is not the quotient rule: we are comparing $f'/g'$ , not calculating $(f/g)'$ . Recheck the hypotheses before repeating the operation rather than continuing until a convenient expression appears.

Worked example: transform a product

For $x\ln x$ as $x\to0^+$ , the form is $0\cdot(-\infty)$ , so the theorem does not apply directly. Rewrite it as $\ln x/(1/x)$ , an infinite-over-infinite form. The derivative ratio is $(1/x)/(-1/x^2)=-x\to0$ . Therefore the original product tends to zero from below.

For a positive-base variable power, take logarithms first. If $y=x^x$ , then $\ln y=x\ln x\to0$ , and continuity of the exponential gives $x^x\to1$ as $x\to0^+$ . The transformation must respect the real logarithm's positive domain.

Why the checks cannot be skipped

The quotient $(1+x)/x$ as $x\to0^+$ tends to $+\infty$ , not $1$ . Differentiating numerator and denominator would incorrectly suggest $1$ because the original form is $1/0$ , not an allowed indeterminate form.

Likewise, a limit can exist even if the derivative ratio oscillates. L’Hôpital's theorem is a sufficient route to a conclusion, not a test that every existing quotient limit must pass. Algebra, squeezing and growth comparisons remain valuable alternatives and can be shorter.

Practice

Evaluate $\lim_{x\to0}\ln(1+x)/x$ .
Evaluate $\lim_{x\to\infty}\ln x/x$ .
Explain why differentiating numerator and denominator in $\lim_{x\to0}(2+x)/(3+x)$ is invalid as an application of this theorem.

Show worked solutions

This is $0/0$ and the derivative ratio is $1/(1+x)\to1$ .
This is infinite over infinite and the derivative ratio is $1/x\to0$ .
The original limit is $2/3$ by continuity; its form is not indeterminate. A derivative ratio of $1$ is irrelevant.