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$$\lim _{x \rightarrow 0} \left(\frac{ \sin x}{x}\right)^{1/x}$$

I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.

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This reduces to $\lim \exp(\frac{(\frac{sin(x)}{x})'}{\frac{sin(x)}{x}}) = \exp(\lim \frac{x\cdot \cot(x)- 1}{x})$. –  Chris K Dec 13 '13 at 6:47
    
$f(x)^{g(x)}=e^{g(x)\ln{f(x)}}$ –  Poppy Dec 13 '13 at 6:51
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You cannot "solve" a limit. You may evaluate a limit. Why not try using words according to their standard usages? –  Michael Hardy Dec 13 '13 at 6:53
    
@Poppy.So, according to you, $\lim_{x\to 0}(\dfrac{sinx}{x})^{\frac{1}{x}}$=$e^{\frac{1}{x}ln{\frac{sinx}{x}}}$.Now, to me evaluating that is also a challenge! Please elaborate your solution! –  Hawk Dec 13 '13 at 7:06

6 Answers 6

up vote 3 down vote accepted

Hint : Take its logarithm, use the fact that $\ln a^b=b\ln a=\dfrac{\ln a}{1/b}$, and apply l'Hopital $3$ or $4$ times.


$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$

$$=\lim_{x\to0}\frac{(1\cdot\cos x-x\cdot\sin x)-\cos x}{1\cdot\sin x+x\cdot\cos x}=-\lim_{x\to0}\frac{x\cdot\sin x}{\sin x+x\cdot\cos x}=$$

$$=-\lim_{x\to0}\frac{1\cdot\sin x+x\cdot\cos x}{\cos x+(1\cdot\cos x-x\cdot\sin x)}=-\lim_{x\to0}\frac{\sin x+x\cdot\cos x}{2\cdot\cos x-x\cdot\sin x}=-\frac{0+0\cdot0}{2\cdot1-0\cdot0}=-\frac02$$

$$=0\iff L=e^0=1.$$

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I already tried that, it doesn't work. –  dfg Dec 13 '13 at 13:08
    
@dfg: It does work, I did it myself, the answer (after $3$ or $4$ tries) is $0$, and $e^0=1$. –  Lucian Dec 13 '13 at 13:10
    
Could you please post the full solution? Because I tried it multiple times, but got nowhere. –  dfg Dec 13 '13 at 13:11
    
@dfg: Done! :-) –  Lucian Dec 13 '13 at 13:28
    
Awesome, thanks a lot! :D –  dfg Dec 13 '13 at 13:32

For small values of $x$, the Taylor series for $\sin(x)$ is $x - x^3 / 6$; then, for $\sin(x)/ x$, the expansion around $x=0$ is $1 - x^2 /6$. Now, remember that, for small values of $y$, $(1+y)^a$ is approximated by $1 + a y$ (another Taylor series). So, for your expression, you arrive to

$$(1 - x^2 /6)^{1/x}$$

which is almost $(1 - x / 6)$. So, your limit is $1$.

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@Mhenni Benghorbal. Thanks for editing. –  Claude Leibovici Dec 13 '13 at 7:51
    
Now problem. You are welcome. –  Mhenni Benghorbal Dec 13 '13 at 7:52

Let $$ y = \frac{\sin x}{x}-1 $$ and notice that for $x\to 0$ also $y\to 0$. Then remember that $$ \lim_{y\to 0} \left(1+y\right)^{\frac 1 y} = e $$ while (use Hopital or Taylor here) $$ \frac{y}{x} = \frac{\frac{\sin x}{x} - 1}{x} = \frac{\sin x - x}{x^2} \to 0. $$ So your limit is $$ (1+y)^\frac 1x = \left((1+y)^\frac{1}{y}\right)^{\frac{y}{x}} \to e^0=1 $$

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Notice by Young's Inequality since $(\frac{1}{x} + (1 - \frac{1}{x})) = 1$, then

$$ (\frac{\sin x}{x} )^{1/x}=(\frac{\sin x}{x} )^{1/x} 1^{1 - \frac{1}{x}} \leq \frac{\sin x}{x^2} + 1 - \frac{1}{x} = \frac{\sin x - x}{x^2} + 1$$

Now, for positive $x$ and for $a \leq 1$, we have that $a^x \leq x +1 $. Hence $a^{1/x} \geq \frac{1}{x+1}$. Now, since $\frac{\sin x}{x} \leq 1$, we apply this inequality with $a = \frac{\sin x}{x} $ to obtain

$$ (\frac{\sin x}{x} )^{1/x} \geq \frac{1}{x+1}$$. Hence we have

$$ \frac{1}{x+1} \leq (\frac{\sin x}{x} )^{1/x} \leq \frac{\sin x - x}{x^2} + 1$$.

Now, since $\lim{ \frac{1}{x+1} } = 1 $ and

$$ \lim (\frac{\sin x - x}{x^2} + 1 ) = \lim ( \frac{\sin x - x}{x^2} ) = 1 + \lim ( \frac{\cos x - 1}{2x} ) = 1 + \lim ( \frac{- \sin x}{2} ) = 1 $$

Now, result follows by the squeeze trick.

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$\lim\limits_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{x}}=\lim\limits_{x\to0}e^{\frac{\ln{\frac{\sin{x}}{x}}}{x}}=$ $e^{\lim\limits_{x\to0}{\frac{\ln{\frac{\sin{x}}{x}}}{x}}}=\frac{0}{0}$ $=e^{\lim\limits_{x\to0}\frac{x}{\sin{x}}\frac{x\cos{x}-\sin{x}}{x^2}}$, so you were right, it doesn't simplify nicely. I thought it would, sorry. If you are familiar with ~ and o, you can finish this using $\sin{x}$~$x-\frac{x^3}{6},x\to0$.

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A suggestion to showing work for this problem would be to show that lim x>0 of sinx/x=1 and then just simply applying the definition of the limit.(you will end up with 1^(1/x) and since 1 is unitary 1 to any power is just 1.

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It wouldn't work like this. $(1+x)^{1/x}\to e$ as $x\to 0$, yet $1+x\to 0$. –  TZakrevskiy Dec 13 '13 at 15:22

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