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I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$

I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$

So it's enough to find: $$\lim_{x\to 0^+}\frac{\ln(\frac {\sin x}{x})}{x}$$

I tried to use L'Hôpital's rule but it got me nowhere. Any ideas?

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    $\begingroup$ This might help math.stackexchange.com/questions/605202/… $\endgroup$
    – shredalert
    Feb 3, 2017 at 11:36
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    $\begingroup$ Thanks, this actually solved my problem. Apperantly I had to use L'Hôpital 3 times, stopped after 2. $\endgroup$
    – user401516
    Feb 3, 2017 at 11:58
  • $\begingroup$ properties of ln on ln(sinx / x) = ln(sin x) - ln(x) - I think you can then use l'hopital 3 times to show ln(f(x)) = $-\infty$ - and get the result of zero for the limit - it ends up based around -csc x - edit you did too, I thought I was going mad actually! $\endgroup$
    – Cato
    Feb 3, 2017 at 12:17

5 Answers 5

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Hints:

  1. Write $$\left(\frac{\sin x }{x} \right)^\frac{1}{x} = \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right).$$
  2. $\exp$ is continuous, $\lim_{x\to 0} \frac{\sin x }{x}=1$ and $\ln 1=0$.
  3. Evaluate $\lim_{x\to 0}\frac{1}{x} \ln \left(\frac{\sin x }{x} \right)$ by using L'Hospital.

For step 3 use $$\ln \left(\frac{\sin x }{x} \right)' =\frac{\cos x}{\sin x} - \frac{1}{x} = \frac{x\cos x - \sin x}{x \sin x}.$$ For $x\to 0+$ we have an expression like $0/0$ again. Now apply L'Hospital again:

$$\lim_{x\to 0+}\frac{x\cos x-\sin x}{x\sin x}=\lim_{x\to 0+}\frac{(\cos x-x\sin x)-\cos x}{\sin x+x\cos x}=-\lim_{x\to 0+}\frac{x\sin x}{\sin x+x\cos x}$$

$$=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{\cos x+(\cos x-x\sin x)}=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{2\cos x-x\sin x}=0.$$

So we get $$ \lim_{x\to 0+} \left(\frac{\sin x }{x} \right)^\frac{1}{x}= \lim_{x\to 0+} \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right) = \exp \left( \lim_{x\to 0+} \ln \left(\frac{\sin x }{x}\right) \right) = \exp 0 =1.$$

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  • $\begingroup$ part 3 is exactly where I got stuck $\endgroup$
    – user401516
    Feb 3, 2017 at 11:52
  • $\begingroup$ Those steps are all in the question already, and the answer stops righ at the end of the question $\endgroup$
    – tylo
    Feb 3, 2017 at 11:55
  • $\begingroup$ Yes, I know. I accidentally deleted my last part. $\endgroup$
    – user339727
    Feb 3, 2017 at 12:11
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Using Taylor polynomials, $$ \frac1x\,\log\frac {\sin x}x=\frac1x\,\log (1-O (x^2))=O (x)\to0. $$

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Using the big-O notation in its common sense, $$\sin x = x + O(x^3),$$ whence $$\ln(\frac {\sin x}x)=\ln\frac{x+O(x^3)}x=\ln(1+O(x^2))=O(x^2).$$

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Since $f(x)=e^x$ is a continuous function, we obtain: $$\lim_{x\to 0^+}\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x}}=\lim_{x\to 0^+}\left(1+\frac{\sin{x}}{x}-1\right)^{\frac{1}{\frac{\sin{x}}{x}-1}\frac{\frac{\sin{x}}{x}-1}{x}}=e^{\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x^2}}=1$$ Because $\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x^2}=\lim\limits_{x\rightarrow0}\frac{\cos{x}-1}{2x}=0$

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There is an interesting massive overkill: we may prove that in a right neighbourhood of the origin we have $$1-\frac{x^2}{6}\leq\frac{\sin x}{x}\leq \exp\left(-\frac{x^2}{6}\right) \tag{1}$$ hence the wanted limit is trivially $\exp(0)=1$. By the Weierstrass product for the sine function: $$ \log\left(\frac{\sin x}{x}\right) = \sum_{n\geq 1}\log\left(1-\frac{x^2}{\pi^2 n^2}\right)\leq\sum_{n\geq 1}-\frac{x^2}{\pi^2 n^2}=-\frac{\zeta(2)}{\pi^2}x^2=-\frac{x^2}{6}\tag{2} $$ and we are done, since the inequality on the left of $(1)$ just follows from the Taylor series of $\text{sinc}(x)$.

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  • $\begingroup$ That is a very nice overkill. (+1) $\endgroup$
    – user339727
    Feb 3, 2017 at 12:21

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