1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you use Taylor series? $\endgroup$ – Matthew Leingang Jul 10 '17 at 16:37
  • 3
    $\begingroup$ 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
3
$\begingroup$

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

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} $$\implies \mathrm L=-\frac{1}{6}$$

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.