2

Assume that you want to evaluate $$\lim_{x \to 0} \frac{x-\sin x}{x^3}.$$ Of course, you may apply $\textbf{de l'Hôpital's Rule}$, that's natural and simple. But what if you are required to avoid using the more advanced knowledge such as “de l'Hôpital”，“Taylor” and so on？Now,someone gives a solution for the problem as follows.

Since$$\begin{align*}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}&=\lim\limits_{x \to 0} \dfrac{2x-\sin 2x}{(2x)^3}\\[4pt]&=\lim\limits_{x \to 0} \dfrac{2x-2\sin x\cos x}{(2x)^3}\\[4pt]&=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x+\sin x-\sin x\cos x}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x(1-\cos x)}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x}{x}\cdot\lim_{x \to 0}\frac{1-\cos x}{x^2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\cdot 1\cdot \frac{1}{2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{8},\end{align*}$$ which sets up an equation on the limit, hence, by solving the equation, we obtain that$$\lim_{x \to 0} \frac{x-\sin x}{x^3}=\frac{1}{6}.$$ What about the solution? Is it rigorous? Hope to read your comments. Thank you!

Blue
  • 75,673
mengdie1982
  • 13,840
  • 1
  • 14
  • 39
  • 2
    If you know the limit exists, this shows it is $1/6$. – Angina Seng May 19 '18 at 14:04
  • 2
    Yes,but he didn't prove the existence of that limit. – mengdie1982 May 19 '18 at 14:06
  • For methods which don't assume that the limit exists, see here: https://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3 – Hans Lundmark May 19 '18 at 14:43
  • See https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee May 19 '18 at 16:15
  • To be more precise, this argument shows that, if you know that the limit exists and is finite, then that limit is $\frac16$. Both $L=\infty$ and $L=-\infty$ "satisfy" the equation $L=\frac14 L + \frac18$. – Blue May 21 '18 at 05:04

1 Answers1

1

There's a typo: when you wrote (after the third equality) $\sin\cos x$, you meant $\sin x\cos x$.

This proves correctly that if the limit exists then it must be equal to $\frac16$, but that's all.

José Carlos Santos
  • 427,504
  • Thanks! I have corrected that! – mengdie1982 May 19 '18 at 14:11
  • But it's exactly the matter. He didn't show that the limit exists. And if he wants to show the existence, he need to know the limit value. – mengdie1982 May 19 '18 at 14:18
  • @mengdie1982 Yes, he didn't show that the limit exists. However, I disegree with the assertion that “if he wants to show the existence, he need to know the limit value”. Often, one can prove that a limit exists without computing it. – José Carlos Santos May 19 '18 at 14:22
  • Sir,I understand your opinion. For example, we can show that $\lim\limits_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n$ exsits without evaluating the precise value. But for the problem I posted above, is there the similar method for proving the existence? – mengdie1982 May 19 '18 at 14:28
  • @mengdie1982 For instance, it can be proved that if a function $f$ is of class $C^3$, and if $f(0)=f'(0)=f''(0)=0$, then the limit $\lim_{x\to0}\frac{f(x)}{x^3}$ always exists. And then we can apply this to $f(x)=x-\sin(x)$. – José Carlos Santos May 19 '18 at 14:34
  • Thanks! But it's more more... advanced knowledge... – mengdie1982 May 19 '18 at 14:36
  • @mengdie1982 I don't understand what you're after. You've told by me and you've been told in the comments that what your friend did doesn't prove that the limit exists. What else do you want? – José Carlos Santos May 19 '18 at 14:40