Find the limits:

$\lim\limits_{x\rightarrow \infty} \frac{x+\sin x}{x}$

Since the numerator and denominator tends to infinity as $x$ tends to infinity, then applying Lhopital's rule:

$\lim\limits_{x\rightarrow \infty} \frac{x+\sin x}{x} = \lim\limits_{x\rightarrow \infty} \frac{1 + \cos x}{1}$

since $\cos x$ has no limit as $x$ tend to infinity ($\cos x$ oscillates between $-1$ and $1$), I conclude that the limit of $\frac{x + \sin x }{x}$ as $x$ tends to infinity does not exist too.

Why is this answer wrong (the correct answer is 1) and at what point should I have realized that I made a mistake and abort this solution and try something else? Is $\frac{1 + \cos x}{1}$ considered an indeterminate form?

  • 13
    $\begingroup$ L'Hôpital says that if the limit of $f'/g'$ exists, then the limit of $f/g$ exists. If the limit of $f'/g'$ doesn't exist, you can't say anything about the original limit. To find the limit it, write it as the sum of two limits. $\endgroup$
    – Git Gud
    Oct 22 '14 at 18:30
  • 7
    $\begingroup$ This is a pretty good cautionary example about using L'Hospital without care. $\endgroup$
    – Will Jagy
    Oct 22 '14 at 18:34

L'Hospital rule says that if $\lim_{x \to \infty} f(x)=\lim_{x \to \infty}g(x)= \infty$ and $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ exists then $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ exists and is the same thing.

L'Hospital doesn't say anything about what happens if $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ doesn't exists, and the converse of L'H is not true.

To understand why this happens, you have to look at the proof of L'H. The proof of L'H uses the mean value theorem to deduce that there exists some $c_x$ which depends on $x$ such that $c_x \to \infty$ and $$\frac{f(x)}{g(x)}= \frac{f'(c_x)}{g'(c_x)}$$

Now, $\lim_x \frac{f'(x)}{g'(x)}$ calculates the limit using ALL $x$, while $\lim_x\frac{f'(c_x)}{g'(c_x)}$ calculates the limit using only SOME of the $x$. If the limit exists for all $x$, then it does exists for some of the $x$, but the other way around it is not true.

In your example, what happens most probably is that while $$\lim\limits_{x\rightarrow \infty} \frac{1 + \cos x}{1}$$ does not exist, when you apply the MVT you get $$\frac{x+\sin x}{x} =\frac{1 + \cos c_x}{1}$$ and each $c_x$ is very very close to some $\frac{\pi}{2}+2k \pi$. Moreover, when $x$ goes to infinity the approximation $c_x \sim \frac{\pi}{2}+2k \pi$ becomes better and better.

  • $\begingroup$ A nice solution $\endgroup$
    – haidangel
    Feb 28 at 8:39

$\lim_{x\rightarrow \infty} \frac{x + \sin(x)}{x}$

notice for $x\geq 1$ you have this relation:

$\frac{x - 1}{x} \leq \frac{x + \sin(x)}{x} \leq \frac{x + 1}{x}$


$1 = \lim_{x\rightarrow \infty} \frac{x - 1}{x} \leq \lim_{x\rightarrow \infty} \frac{x + \sin(x)}{x} \leq \lim_{x\rightarrow \infty} \frac{x + 1}{x} = 1$

As a result, the limit of the original function is 1 by the pinching theorem.


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