9

This is how I would evaluate $\lim\limits_{x \to \infty} \dfrac{x+ \sin x}{x+ 2 \sin x}$

$=\lim\limits_{x \to \infty} \dfrac{x \left( 1+ \frac{\sin x}{x} \right)}{x \left(1+ 2 \cdot \frac{ \sin x}{x} \right)}$

$= \dfrac{1+0}{1+2 \cdot 0} = 1$

But now applying L'hopitals Rule, I get

$\lim\limits_{x \to \infty} \dfrac{1+ \cos x}{1+ 2 \cos x}$

Since $\cos x $ just oscillates between $[-1,1]$ I think we can conclude the limit doesn't exist.

What is going on here?

StubbornAtom
  • 17,052
William
  • 4,893
  • See also here: https://math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently. – Zacky Dec 27 '18 at 13:50
  • This is also explained on the wikipedia page (https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Requirement_that_the_limit_exist) where there is a simpler example such as $(x + \cos x)/x$. – Ben Dec 27 '18 at 13:51
  • See also https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-this-case – Barry Cipra Dec 27 '18 at 14:06
  • See also here – Hirshy May 10 '19 at 07:29

2 Answers2

15

L'Hospital's rule contains an assumption that $\lim_{x \to a} f'(x)/g'(x)$ exists, which is not true in this case.

littleO
  • 51,938
  • 3
    Unfortunately, students often forget the hypotheses for a result when they try to apply it. – GEdgar Dec 27 '18 at 13:51
2

Because your function doesn't satisfy the hypothesis. If you are studing the limit $x\to c$, in order to apply the theorem the function $g=x+2\sin x$ must be differentiable and $g'(x)\ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+\infty) where $g' \ne 0$. But $g'(x)=0$ $\forall x=-\frac{\pi}{4}+2k\pi$, so it doesn't exists a set like that.

ecrin
  • 1,423
  • The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2\sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ... – DonAntonio Dec 27 '18 at 14:06
  • Thank you for the explanation! – ecrin Dec 27 '18 at 14:09