L'Hôpital's Rule

L'Hôpital's rule, also written L'Hopital's rule, is a calculus method for quotients whose limits collapse to $0/0$ or $\infty/\infty$ after direct substitution. It is not a universal limit solver — it is a targeted tool for those two indeterminate forms.

When to reach for it

Start with a quotient:

The rule applies only when all of these hold:

1. Direct substitution gives $0/0$ or $\infty/\infty$.
2. The expression is genuinely a quotient (not $0 \cdot \infty$ or $\infty - \infty$ left unrewritten).
3. $f$ and $g$ are differentiable near $a$, with $g'(x) \ne 0$ nearby.

When they do, the theorem lets you compare derivatives instead of values:

provided the right-hand limit exists or is $+\infty$ or $-\infty$. In early calculus this fits best with exponentials, logarithms, and trig functions near special points, with quotients that still look indeterminate after substitution, and with growth-rate comparisons such as polynomial versus exponential.

The steps, in order

1. Substitute first. Plug in the target value and confirm you actually get $0/0$ or $\infty/\infty$. An indeterminate form is the entry ticket.
2. Confirm it is a quotient. If the expression is $0 \cdot \infty$ or $\infty - \infty$, rewrite it as a single fraction before going further.
3. Differentiate top and bottom separately. Take $f'(x)$ and $g'(x)$ — this is not the quotient rule, just two independent derivatives.
4. Re-evaluate the new limit. Substitute again into $\frac{f'(x)}{g'(x)}$.
5. Repeat or stop. If you still get an indeterminate form and the conditions still hold, apply the rule again; otherwise you are done.

Why the derivative step works

When both parts go to $0$, or both grow without bound, their raw values say nothing decisive. The derivative step compares how fast the numerator and denominator are changing near the point, and that ratio of rates is what the limit was really asking about. That is why one differentiation often turns a murky limit into one you can read at a glance.

Full worked example: $\lim_{x \to 0} \frac{e^x - 1}{x}$

Direct substitution gives

so the form is eligible and it is already a quotient. Differentiate top and bottom:

The new limit is

So

One derivative step settled it — the ideal case for this method.

Where each step traps people, plus a self-check

The first step traps people who skip the form check and apply the rule to every hard limit. The second traps those who differentiate $0 \cdot \infty$ or $\infty - \infty$ directly instead of rewriting first. The third invites errors when factoring, rationalizing, or a known limit would have been cleaner. And many forget the conditions entirely — an indeterminate form alone is not the full theorem.

Before applying the rule, run through this self-check: Does substitution give $0/0$ or $\infty/\infty$? Is it written as a quotient? Are numerator and denominator differentiable near the point? Does differentiating make the limit easier rather than messier? If any answer is no, pause and simplify or pick another approach.

For your own pass, try

Direct substitution gives $0/0$, and one derivative step turns it into a basic limit. As a cross-check, redo it with the approximation $\ln x \approx x - 1$ near $x = 1$ and confirm both methods agree.