MISSION · Calculus
L'Hôpital's Rule
When substitution gives nonsense, differentiate the top and bottom separately
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The idea in one sentence
When a limit comes out as \(\frac{0}{0}\), differentiate the top and the bottom separately and try again — the new fraction has the same limit.
The problem it solves
Substitute \(x = 0\) into \(\dfrac{\sin x}{x}\) and you get \(\dfrac{0}{0}\). That is not an answer. It is not zero, it is not one, it is not undefined-so-no-limit — it is a statement that substitution has told you nothing, because two competing effects are cancelling and you cannot see which wins.
Such a form is called indeterminate, and the whole question is which of the top and bottom is racing to zero faster.
The rule
If \(\lim \frac{f(x)}{g(x)}\) gives \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\), then
\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]provided that second limit exists.
For \(\dfrac{\sin x}{x}\): differentiate top and bottom to get \(\dfrac{\cos x}{1}\), which at \(x = 0\) is \(1\). So the limit is 1.
For \(\dfrac{e^x - 1}{x}\): substitution gives \(\frac{0}{0}\), differentiating gives \(\dfrac{e^x}{1} \to 1\).
This is not the quotient rule
Say it out loud once: you are not differentiating \(\frac{f}{g}\). You are replacing it with \(\frac{f'}{g'}\), a completely different expression, and there is no quotient rule anywhere.
The reason it works is that near \(a\), both functions are approximately their tangent lines: \(f(x) \approx f'(a)(x - a)\) and \(g(x) \approx g'(a)(x - a)\). Divide one by the other and the \((x-a)\) cancels, leaving \(\frac{f'(a)}{g'(a)}\). Racing to zero at rates \(f'(a)\) and \(g'(a)\), the ratio of the rates is what survives.
Checking the form first, every time
The rule is only valid on an indeterminate form, and applying it to an ordinary limit produces a confident wrong answer with no warning. Consider \(\lim_{x \to 0} \frac{x + 2}{x + 1}\): substitution gives \(2\), but differentiating top and bottom gives \(\frac{1}{1} = 1\). The rule was never licensed, and it lied.
So the order is fixed: substitute, look at what you got, and only then reach for the rule.
Repeating, and the other forms
If the new fraction is still \(\frac{0}{0}\), apply it again. \(\lim_{x \to 0} \frac{1 - \cos x}{x^2}\) goes to \(\frac{\sin x}{2x}\), still \(\frac{0}{0}\), then to \(\frac{\cos x}{2} = \frac{1}{2}\).
Five other indeterminate forms exist — \(0 \cdot \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), \(\infty^0\) — and none can be attacked directly. Each has to be rewritten as a quotient first, usually by combining into one fraction, or by taking logs for the three with exponents.
The forms that are not indeterminate need no rule at all. \(\frac{5}{0^+}\) is \(+\infty\); \(\frac{0}{5}\) is \(0\). Only the genuine standoffs qualify.
Common mix-ups
- Using the quotient rule by mistake. Top and bottom are differentiated separately and independently.
- Applying it to a determinate form. Always substitute first.
- Forgetting to re-check after each pass. Once the form stops being indeterminate, stop and substitute.
- Looping forever. \(\frac{e^x}{e^x}\) and \(\frac{\sqrt{x^2+1}}{x}\) regenerate themselves; simplify algebraically instead.
- Circular reasoning with \(\frac{\sin x}{x}\). Using the rule here is fine for exam purposes, but the derivative of \(\sin x\) is itself proved from this limit, so it is not a proof.
Check yourself
Evaluate \(\lim_{x\to 0}\frac{\sin x}{x}\) and \(\lim_{x\to 0}\frac{e^x - 1}{x}\), and say what \(\frac{f}{g}\) gets replaced by.
Both substitute to \(\frac{0}{0}\). Differentiating separately gives \(\frac{\cos x}{1} \to 1\) and \(\frac{e^x}{1} \to 1\). The replacement is \(\frac{f'}{g'}\) — never \(\left(\frac{f}{g}\right)'\).
LEVEL 3 · REFLECT
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