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I'm trying to calculate the following limit:

$$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$

What I did is writing it as:

$${e^{\frac{1}{x}\ln \left( {\frac{{\sin x}}{x}} \right)}}$$

Therefore, we need to calculate:

$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x}$$

Now, we can apply L'Hopital rule, Which I did:
$$\Rightarrow cot(x) - {1 \over x}$$

But in order to reach the final limit two more application of LHR are needed. Is there a better way?

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4 Answers 4

up vote 7 down vote accepted

Note that as $x\to 0^+$ you have $$\frac{\ln\frac{\sin x}x}x\sim \frac{\frac{\sin x}x-1}x\sim \frac{\sin x-x}{x^2}\sim \frac{-x^3}{6x^2}\to 0$$.

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How can I explain those steps? I understand what you did, but how can I write it down formally? certainly, those tildas won't do the work :) –  SuperStamp Jan 11 at 23:29
    
The tilda means asmiphtotic (see en.wikipedia.org/wiki/Asymptotic_analysis), so my writing is formally complete and correct. In particular, I applied $\ln(1+x)\sim x$ and $\sin(x)-x\sim -\frac{x^3}6$ as $x\to 0$. –  Fabio Lucchini Jan 11 at 23:31
    
hmm..OK, didn't know that. Thanks! –  SuperStamp Jan 11 at 23:32
    
Would you add verbal explanation to this in a proof? –  SuperStamp Jan 11 at 23:32
3  
I used an "asymptotic" version of Taylor-Mac Laurin formula for $\sin$ and $\ln$. –  Fabio Lucchini Jan 11 at 23:38

Elementary proof using well known limits and inequalities: $$\frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x} = \frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\frac{{\sin x}}{x}-1 }{x} =\frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\sin x-x }{x^2}$$ but for $x >0$ we have: $x-\frac{x^3}{6} \le \sin x \le x \Rightarrow -\frac{x^3}{6} = x-\frac{x^3}{6}-x\le\sin x -x \le 0 \Rightarrow -\frac{x^2}{6} \le\frac{\sin x-x }{x^2} \le 0$ the rest is squeeze theorem.

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nice one, I don't think I could come up with that. –  SuperStamp Jan 11 at 23:46

According to Bernoulli's Inequality, for $0\le x\le1$ $$ \left(\frac{\sin(x)}{x}\right)^{1/x}=\left(1-\frac{x-\sin(x)}{x}\right)^{1/x}\ge1-\frac1x\frac{x-\sin(x)}{x}\tag{1} $$ This answer provides an elementary proof of Bernoulli's Inequality for rational exponents.

In this answer, it is shown geometrically that for $0\lt x\lt\frac\pi2$, we have $\sin(x)\le x\le\tan(x)$. Therefore, $$ \begin{align} 0\le\frac{x-\sin(x)}{x^2}&\le\frac{\tan(x)-\sin(x)}{\sin^2(x)}\\[4pt] &=\frac{\tan(x)-\sin(x)}{\sin(x)\tan(x)}\sec(x)\\[4pt] &=\frac{1-\cos(x)}{\sin(x)}\sec(x)\\[4pt] &=\frac{\sin(x)}{1+\cos(x)}\sec(x)\\[4pt] &\to\frac02\cdot1\\[9pt] &=0\tag{2} \end{align} $$ Thus, by the Squeeze Theorem, $$ \lim_{x\to0}\frac{x-\sin(x)}{x^2}=0\tag{3} $$ and therefore, $$ 1\ge\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{1/x}\ge1-0\tag{4} $$

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Thank you sir, nice approach! –  SuperStamp Jan 11 at 23:53
    
@SuperStamp: I've just removed L'Hospital :-) –  robjohn Jan 12 at 0:16
    
Why did you remove LHR? –  SuperStamp Jan 12 at 0:18
2  
To make the limit more elementary. I thought it would be nice to use only pre-calculus concepts. –  robjohn Jan 12 at 0:21
    
Cool. very appreciated! :) –  SuperStamp Jan 12 at 0:22

Here, as often, Taylor series are much more efficient than L'Hopital's rule.

$$\begin{eqnarray}\dfrac{\sin(x)}x \ &=&\ \color{#c00}1\color{#0a0}{\,-\,\dfrac{x^2}6+O(x^4)}\!\!\!\!\!\!\!\\&=&\ \color{#c00}1\, +\,\color{#0a0}z\\ \Rightarrow\ \ x^{-1}\log\left(\dfrac{\sin(x)}x\right) &=&\ x^{-1}\log(\color{#c00}1+\color{#0a0}z)\ &=&\ x^{-1}\left(\ \ \ \color{#0a0}z\ \ \ +\ \ O(z^2)\right)\\ && &=&\ x^{-1}\left(-\color{#0a0}{\dfrac{x^2}6+O(x^4)} \right)\\ && &=&\ -\dfrac{x}6\, +\ O(x^3)\\ && &&\!\!\!\!\!\!\!{\rm which}\overset{\phantom{I}} \to \ 0 \ \ \, {\rm as}\, \ \ x\to0^+\\ \end{eqnarray}\qquad\quad\qquad\qquad$$

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