I have got this problem on limits $$ \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$$ What I am doing here is that taking log and then applying L-Hospital's Rule $$y= \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2} $$ then $$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$ now it becomes $\frac{0}{0}$ type , but problem is how to take derivative here? when differentiating $\ln\left(\frac{\sin(x)}{x}\right)$ it will become $\frac{x}{\sin(x)}$ then we'll have to differentiate one more time like we do in case of $f(f(x))$ ? $\sin(x)$ and $x$ will be separately differentiated or together ?


$$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$
Becomes: $$\frac{x\cos(x)-\sin(x)}{2x^2\sin(x)}$$ After applying L'Hôpital's rule again differentiating two more times w.r.t $x$, we get:
$$\lim_{x \rightarrow 0} \frac{-\cos(x)}{-2x\sin(x)+2\cos(x)+4\cos(x)}=-1/6$$

  • $\begingroup$ okay that means we have to differentiate taking $\sin(x)/x$ as a single function $\endgroup$ – Tesla Mar 5 '14 at 8:07

Use the Taylor series to get $$\lim_{x\to0}\frac1{x^2}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln\left(\frac{ x-x^3/6+o(x^3)}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln(1-x^2/6+o(x^2))\\=\lim_{x\to0}\frac1{x^2}(-x^2/6+o(x^2))=-\frac16$$ hence the desired limit is $e^{-\frac16}$.

  • $\begingroup$ I wonder why answers who use hopital have often more upvotes than the ones using Taylor approximation, which is so much more elegant.. $\endgroup$ – Ant Mar 5 '14 at 8:26
  • $\begingroup$ @Ant maybe because Taylor expansion is more advanced than L'Hôpital rule (i.e. learned later). $\endgroup$ – Ruslan Mar 5 '14 at 8:33
  • $\begingroup$ @Sami Ben Romdhane Not to be pedantic, but isn't it instead Taylor formula? $\endgroup$ – alex Mar 5 '14 at 8:48
  • $\begingroup$ @alex isn't Taylor approximation the same as Taylor formula? Ant didn't say it's Taylor series. $\endgroup$ – Ruslan Mar 5 '14 at 8:50
  • $\begingroup$ @Ruslan I addressed my comment to Ant. But now I see I made a mistake. Anyway not to you. $\endgroup$ – alex Mar 5 '14 at 8:52

$$\lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}=\lim_{x\rightarrow 0}\left(1+\frac{\sin x}{x}-1\right)^\frac{1}{x^2}=\lim_{x\rightarrow 0}\left[\left(1+\frac{\sin x-x}{x}\right)^\frac{x}{\sin x-x}\right]^{\frac{\sin x-x}{x}\frac{1}{x^2}}=e^{-\frac 1 6}$$

  • $\begingroup$ Rightmost $=$ transition is completely unobvious to me. How did $6$ appear there? $\endgroup$ – Ruslan Mar 5 '14 at 8:35
  • $\begingroup$ @Ruslan $\frac{senx-x}{x^3}=\frac{x-\frac{1}{6}x^3+o(x^3)-x}{x^3}$ $\endgroup$ – alex Mar 5 '14 at 8:43
  • $\begingroup$ Then you just overcomplicate things. First, Taylor approximation is an essential step here. Also, $\frac{\sin x-x}x$ should also have been justified to be similar to $\frac1x$. And, Taylor approximation could be used with more ease as in Sami's answer — without nested exponentiation. $\endgroup$ – Ruslan Mar 5 '14 at 8:49

Using $$\displaystyle \sin (\pi z) = \pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)\;,$$ Then Put $\displaystyle \pi z= x\Rightarrow z=\frac{x}{\pi}$

So $$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \lim_{x\rightarrow 0}\prod_{n=1}^{\infty}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}}$$

Now we can Interchange Product and Limit, We get

$$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}\lim_{x\rightarrow 0}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}e^{-\frac{1}{\pi^2n^2}} = e^{-\sum_{n=1}^{\infty}\frac{1}{\pi^2n^2}} = e^{-\frac{1}{6}}$$


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.