L'Hopital's rule, why is this incorrect?

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

• en.wikipedia.org/wiki/L'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?
– user2093
Mar 29 '12 at 9:16

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.

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.

Read the following Wikipedia article carefully :

L'Hôpital's rule