Find the limit $$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}$$

I found the limit which is $-\frac{1}{6}$ by using L'Hopital Rule. Is there another way to solve it without using the rule? Thanks in advance.

  • $\begingroup$ Can you use Taylor series? $\endgroup$ – Matthew Leingang Jul 10 '17 at 16:37
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    $\begingroup$ Since $\lim_{x\to 0}\frac{\tan x}{x}=1$, you have $\lim_{x\to 0}\frac{x^3}{\tan^3 x}=1$. So you want to compute: $$\lim_{x\to 0}\frac{\sin x - x}{x^3}.$$ Not sure if that makes it easier. (Certainly makes the L'Hopital calculation easier.) $\endgroup$ – Thomas Andrews Jul 10 '17 at 16:44
  • $\begingroup$ You can use Taylor expression of sine function . $\endgroup$ – S.H.W Jul 10 '17 at 16:49
  • $\begingroup$ @ThomasAndrews: math.stackexchange.com/questions/157903/… $\endgroup$ – Hans Lundmark Jul 10 '17 at 16:56
  • $\begingroup$ Note $$ \frac{\sin x - x}{x^3} = \frac{-\frac{1}{6}x^3 + O(x^5)}{x^3}$$ $\endgroup$ – MathematicsStudent1122 Jul 10 '17 at 17:00

$$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}= \lim_{x\rightarrow 0} \frac{\sin x-x}{x^3} \underbrace{\left(\frac {x^3}{\tan^3 x}\right)}_{=1}$$

Now, let $$\mathrm L= \lim_{x\rightarrow 0} \frac{\sin x-x}{x^3}$$

Let $x=3y$, Since $x\to 0 \implies y \to 0$.

$$ \mathrm L= \lim_{y\rightarrow 0} \frac{\sin (3y)-3y}{(3y)^3}$$

\begin{align} \mathrm L &= \lim_{y\rightarrow 0} \frac{3\sin y-4\sin^3 y-3y}{27y^3}\\ &= \lim_{y\rightarrow 0}\frac{3}{27} \left(\frac{\sin y-y}{y^3}\right) -\frac{4}{27} \lim_{y\rightarrow 0}\left(\frac{\sin y}{y}\right)^3\\ &=\frac{3}{27} \cdot \mathrm L-\frac{4}{27} \end{align} $$\implies \mathrm L=-\frac{1}{6}$$

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    $\begingroup$ But that only shows that $L=-\frac{1}{6}$ if $L$ exists. (I'm not the downvoter, though. Not sure why anybody would downvote.) $\endgroup$ – Thomas Andrews Jul 10 '17 at 17:06
  • $\begingroup$ But if you want to use Taylor, this approach isn't needed. By this approach, the answerer hasn't shown $L$ exists - it could be infinite, for example, and $L=\frac{3}{27}L-\frac{4}{27}$ is still true. @YvesDaoust $\endgroup$ – Thomas Andrews Jul 10 '17 at 17:10
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    $\begingroup$ Hmmm, not bad, but strictly speaking, you only proved that if the limit you denoted by $L$ exists, it must be $-1/6$. You did not prove that it exists. $\endgroup$ – Professor Vector Jul 10 '17 at 17:12
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    $\begingroup$ I didn't say $L=-4/27$, I said $-4$ should be $-\frac{4}{27}$. @YvesDaoust I was not correcting the answer, but an intermediate equation. $\endgroup$ – Thomas Andrews Jul 10 '17 at 17:13
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    $\begingroup$ I won't downvote, but this is wrong. Unless you prove the limit exists this is like proving that $1+2+4+\dots+2^n+\dotsb=-1$, which can be done with a similar algebraic manipulation. $\endgroup$ – egreg Jul 10 '17 at 17:54

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