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Any suggestions for evaluating the limit $$\lim_{x\to0}(x\tan x)^x$$

I have tried writing $\tan$ as $\dfrac{\sin}{\cos}$ and then got the Taylor series of them but it didn't lead me somewhere. Thanks a lot

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Assuming the limit exists and is equal to $L$, take logs:

$$\log{L} = \lim_{x \to 0} x \log{(x \tan{x})}$$

Use $\tan{x} \sim x$ in this limit. Then use

$$\lim_{y \to 0} y \log{y} = 0$$

and the limit should come out easily:

$$\log{L} = \lim_{x \to 0} x \log{x^2} = \lim_{x \to 0} 2 x \log{x} = 0$$

Therefore, $L=1$.

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  • $\begingroup$ Thanks a lot! Only question I got is why is there tanx ~ x? $\endgroup$ – darkchampionz Jul 18 '13 at 16:30
  • $\begingroup$ Because $\tan{x} = \sin{x}/\cos{x}$ and $\cos{x}$ is near $1$ and $x$ is small, while $\lim_{x \to 0} \sin{x}/x = 1$. $\endgroup$ – Ron Gordon Jul 18 '13 at 16:31
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$$\large(x\tan x)^x=e^{\log(x\tan x)^x}=e^{x\log(x\tan x)}$$

Then $$\large\lim_{x\to 0}(x\tan x)^x=e^{\lim_{x\to 0}x\log(x\tan x)}\tag{since $\exp(x)$ is continuous}$$

Now $$\lim_{x\to 0}x\log(x\tan x)=\lim_{x\to 0}\dfrac{\log x+\log\tan x}{\dfrac{1}{x}}$$

Apply L'Hopital's rule.

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The limit exists if and only if $$ \lim_{x\to0} \log((x\tan x)^x)=\lim_{x\to0}(x\log(x\sin x)-x\log\cos x) $$ exists, since the exponential is a continuous (increasing) function. The second summand gives no problem, because its limit is clearly $0$. So we compute \begin{align} \lim_{x\to0}x\log(x\sin x)&= \lim_{x\to0}x\log\left(x^2\frac{\sin x}{x}\right)\\ &=\lim_{x\to0}2x\log x +\lim_{x\to0}x\log\left(\frac{\sin x}{x}\right). \end{align} Again, the second summand has limit $0$ and also the first one is well-known to have limit $0$. Therefore $$ \lim_{x\to0} \log((x\tan x)^x)=0 $$ and so $$ \lim_{x\to0} (x\tan x)^x)=e^0=1. $$

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Hint:let $y=(x\tan x)^x$ , use $ x\sim \tan x$ and $\lim_{x\to0}x\ln x=0$ $$\large{\lim_{x\to0}y=\lim_{x\to0}e^{\ln y}=\lim_{x\to0}e^{x\ln(x\tan x)}=\lim_{x\to0}e^{x\ln(x^2)}=\lim_{x\to0}e^{2x\ln(x)}=e^{0}=1}$$

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