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As the title says, prove that $\lim_{x\to 0}\big(\frac{x}{\sin x}\big)^{1/x^2} = e^{\frac{1}{6}}$. I've tried using L'Hoptial's rule like so:

If $\lim_{x\to 0}\big(\frac{x}{\sin x}\big)^{1/x^2} = L$, then $\log L = \lim_{x\to 0}\frac{\log(\frac{x}{\sin x})}{x^2}$. Since this is of the form $\frac{0}{0}$, we can apply L'Hopital's rule to get $\log L=\lim_{x\to 0}\frac{\frac{1}{x}-\cot (x)}{2x}$. Since this is also of the form $\frac{0}{0}$, we can apply L'Hopital's rule again to get $\log L = \lim_{x\to 0}\frac{\csc^2(x)-\frac{1}{x^2}}{2}$.

I'm not sure where to go from here, or if I'm really headed in the right direction. I'm also not sure how to show that the intermediate steps are $\frac{0}{0}$, but this is less concerning to me. How do I show that this limit is $e^{\frac{1}{6}}$, or that the last limit is $\frac{1/3}{2} = \frac{1}{6}$?

wlyles
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3 Answers3

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After your first use of L'Hopital's Rule, do not immediately apply L'Hopital's Rule again, but do some algebra: $$\frac{\frac1x-\cot x}{2x}=\frac{\sin x-x\cos x}{2x^2\sin x}\ .$$ Then continue with L'Hopital's Rule (and perhaps a bit more algebra at some stage).

David
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  • Ah, I didn't see that. Makes much more sense when you write the top as $\frac{1}{x}-\frac{\cos x}{\sin x}$. After this, it was two more applications of L'Hoptial's, with basically no algebra. Thanks! – wlyles Dec 02 '14 at 04:25
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    Worth remembering: L'H is immensely useful, but not always the best way to go. In fact IMHO pretty much never the first thing to try. – David Dec 02 '14 at 04:33
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You should note that applying the L'Hopital's Rule to $$\lim_{x\to0}\frac{\log(\frac{x}{\sin(x)})}{x^2} = \lim_{x\to0}\frac {1-\frac{x}{\tan(x)}}{x^2} = \frac{1}{2} \lim_{x\to0}\frac {-x + \tan(x)}{x^2 \tan(x)}$$

Then, applying L'Hopital's Rule for some couple times (work not shown), you will get: $$\lim_{x\to0}\frac{\log(\frac{x}{\sin(x)})}{x^2}=\frac{1}{2}\lim_{x\to0} \frac{1}{2 + \cos(2x)}= \frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}$$

Therefore: $$\log L = \frac{1}{6} \implies L= e^{\frac{1}{6}}$$

Joshua
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Without using L'Hopital's rule, as suggested by André Nicolas, you use Taylor series built at $x=0$. Starting with $$\sin(x)= x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\frac{x}{\sin(x)}=1+\frac{x^2}{6}+\frac{7 x^4}{360}+O\left(x^6\right)$$ So, if $$A=\Big(\frac{x}{\sin x}\Big)^{1/x^2}$$ $$\log(A)=\frac{1}{x^2}\log\Big(\frac{x}{\sin x}\Big)$$ Now, using the fact that, for small $y$, $\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$, then $$\log\Big(\frac{x}{\sin(x)}\Big)=\frac{x^2}{6}+\frac{x^4}{180}+O\left(x^6\right)$$ $$\log(A)=\frac{1}{x^2}\log\Big(\frac{x}{\sin(x)}\Big)=\frac{1}{6}+\frac{x^2}{180}+O\left(x^4\right)$$ which shows the limit and how it is approached.

By the way, if you plot in the same graph the two functions $$y=\Big(\frac{x}{\sin x}\Big)^{1/x^2}$$ $$y=\sqrt[6]{e} \left(1+\frac{x^2}{180}\right)$$ uou could amazed to see how close they are to each other at least in the range $-1 \leq x \leq 1$.

Claude Leibovici
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