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As the title says, why is this an incorrect use of L'Hopital's rule? Clearly this isn't the correct limit?$$\lim_{x \to 2} \frac{\sin x}{x^2} =\lim_{x \to 2} \frac{\cos x}{2x} =\lim_{x \to 2} \frac{-\sin x}{2} = \frac{-\sin 2}{2}$$

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2'H%C3%B4pital's_rule. Maybe that as $x\rightarrow 2$, the numerator and the denominator have finite nonzero limits (which, moreover, are different) is the problem? – William Mar 29 '12 at 9:16
up vote 6 down vote accepted

You can use L'Hopital only in the case where nominator and denominator both tend to 0 or both tend to $\infty$. In your case both tend to a finite value so it can be evaluated directly and you don't need to use L'Hopital.

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One way to think about L'Hopital's Rule is that we are replacing the functions with their local linear approximations:

Say $f(x)$ and $g(x)$ are differentiable at $x=a$; then we know that $$\begin{align*} f(x) &\approx f(a) + (x-a)f'(a)&\text{for }x\text{ near }a,\\ g(x) &\approx g(a) + (x-a)g'(a)&\text{for }x\text{ near }a. \end{align*}$$

So, we might expect that $$\frac{f(x)}{g(x)} \approx \frac{f(a)+(x-a)f'(a)}{g(a)+(x-a)g'(a)},\quad\text{for }x\text{ near }a.$$

If $f(a)=g(a)=0$, then the fraction on the right simplifies to $$\frac{f(a)+(x-a)f'(a)}{g(a)+(x-a)g'(a)} = \frac{0+(x-a)f'(a)}{0+(x-a)g'(a)} = \frac{(x-a)f'(a)}{(x-a)g'(a)} = \frac{f'(a)}{g'(a)},$$ which yields the "easy" version of L'Hopital's Rule: if $f(x)$ and $g(x)$ are differentiable at $a$, $f(a)=g(a)=0$, and $g'(a)\neq 0$, then $$\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}.$$

But what if $f(a)\neq 0$ or $g(a)\neq 0$? It should be clear that the simplification given above simply does not work anymore. L'Hopital's Rule has no reason to work then.

(It does work if $\lim\limits_{x\to a}f(x)=\pm\infty$ and $\lim\limits_{x\to a}g(x)=\pm\infty$, but not by the argument above).

In short: it doesn't work because the premises that you need are not present in your example.

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Read the following Wikipedia article carefully :

L'Hôpital's rule

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