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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 ?

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$$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$
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:
$$\lim_{x \rightarrow 0} \frac{-\cos(x)}{-2x\sin(x)+2\cos(x)+4\cos(x)}=-1/6$$

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  • $\begingroup$ okay that means we have to differentiate taking $\sin(x)/x$ as a single function $\endgroup$ – Tesla Mar 5 '14 at 8:07
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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 $e^{-\frac16}$.

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

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  • $\begingroup$ 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
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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

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

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