Calculus – Why Doesn’t L’Hôpital’s Rule Work in This Case?

calculuslimits

I have a very simple question. Suppose I want to evaluate this limit:

$$\lim_{x\to \infty} \frac{x}{x-\sin x}$$

It is easy to evaluate this limit using the Squeeze theorem (the answer is $1$). But here both the numerator and the denominator are going to infinity as $x\to \infty$ so I tried using L'Hospital's rule:
$$\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}$$

However there's no finite $L$ such that $$\lim_{x\to \infty} \frac{1}{1-\cos x}=L$$
which is a contradiction. I don't understand why in this case L'Hopital's rule doesn't work. Both the numerator and the denominator are differentiable everywhere and both are tending to infinity – which is all we need to use this rule.

Best Answer

A precondition in l'Hospital's Rule is that in order for it to apply, the limit $$ \lim_{x\to\infty}\frac{f'(x)}{g'(x)} $$ must exist (but is allowed to be $\pm \infty$). In this case, the limit does not exist, so it does not apply.

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