How do I evaluate this limit :$\displaystyle \lim_{x \to \infty} ({x\sin \frac{1}{x} })^{1-x}$.? I would like to evaluate this limit :$$\displaystyle \lim_{x \to \infty} ({x\sin \frac{1}{x} })^{1-x}$$.
I used taylor expansion at $y=0$ , where $x$ go to $\infty$ i accrossed this 
problem : ${1}^{-\infty }$ then i can't judge if this limit equal's $1$ , 
because it is indeterminate case ,Then is there a mathematical way to 
evaluate this limit ?
Thank you for any help 
 A: I see you're a high school teacher so you're familiar with the following concepts :

$\bullet$ $\sin(\frac{1}{x}) \simeq \frac{1}{x} - \frac{1}{6x^3} \text{  } [\text{as x $\rightarrow$ $\infty$}]$
$\bullet  $ $ \lim_{x \to \infty} (1-\frac{k}{x})^x = e^{-k} $

 
Compile these facts to get :
$$\underset{x \to \infty}{\lim} \bigg(1 - \frac{1}{6x^2} \bigg)^{1-x} = 1$$
A: Compute the limit of the logarithm:
\begin{align}
\lim_{x\to\infty}(1-x)\log(x\sin(1/x))&=
\lim_{t\to0^+}\left(1-\frac{1}{t}\right)\log\frac{\sin t}{t}
\\[6px]
&=\lim_{t\to0^+}\log\frac{\sin t}{t}-\lim_{t\to0^+}\frac{\log\sin t-\log t}{t}\\[6px]
&=-\lim_{t\to0^+}\left(\frac{\cos t}{\sin t}-\frac{1}{t}\right)\\[6px]
&=-\lim_{t\to0^+}\frac{t\cos t-\sin t}{t^2}\cdot
\lim_{t\to0^+}\frac{t}{\sin t}\\[6px]
&=\lim_{t\to0^+}\frac{t\sin t}{2t}\\[6px]
&=0
\end{align}
Of course this can be simplified by recalling that $(\sin t)/t=1+t^2/6+o(t^4)$, so we have
$$
\lim_{t\to0^+}\left(1-\frac{1}{t}\right)\log\left(\frac{\sin t}{t}\right)=
\lim_{t\to0^+}\frac{(t-1)(t^2/6+o(t^4))}{t}=0
$$
A: Let $y=1/x$ and $y\to 0^+$. We have
\begin{align*}
\left(\frac{1}{y}\sin y\right)^{1-1/y} & =\left(\frac{1}{y}\left(y+O\left(y^{3}\right)\right)\right)^{1-1/y}\\
 & =\left(1+O\left(y^{2}\right)\right)^{1-1/y}\\
 & =\exp\left(\ln\left(1+O\left(y^{2}\right)\right)\frac{y-1}{y}\right)\\
 & =\exp\left(O\left(y^{2}\right)\frac{y-1}{y}\right)\\
 & =\exp\left(O\left(y\right)\right).
\end{align*}
So the limit as $y\to0^+$ is 1.
A: I think you have to expand out to the next order term in the Taylor series about infinity.  Thus
$$\left ( x \sin{\frac1{x}} \right )^{1-x}  = \frac{\displaystyle 1-\frac1{6 x^2}}{\displaystyle\left (1-\frac1{6 x^2} \right )^x}$$
Now
$$\left (1-\frac1{6 x^2} \right )^x = \left[ \left (1-\frac1{6 x^2} \right )^{x^2}\right ]^{1/x}$$
so as $x \to \infty$, the expression approaches
$$ \lim_{x \to \infty}\frac{\displaystyle 1-\frac1{6 x^2}}{\displaystyle e^{-1/(6 x)}} = 1$$
A: If you don't want to use Taylor expansion, the following will work.
Firstly, we can get rid of the $1$ in the power, because $$\lim_{x \to \infty} x \sin \left(\frac{1}{x} \right) = 1$$
So we actually want $$\lim_{x \to \infty} \left(x \sin \frac{1}{x} \right)^{-x}$$
By continuity, this is $$\exp \left( - \lim_{x \to \infty} x \log \left(x \sin \frac{1}{x} \right)\right)$$
By the substitution $x = 1/u$, this is $$\exp \left( - \lim_{u \to 0} \frac{1}{u} \log\left(\frac{1}{u} \sin u \right)\right)$$
By L'Hôpital's rule, this is $$\exp \left( - \lim_{u \to 0} \left[ -\frac{1}{u} + \cot u \right]\right)$$
or $$\exp \left( \lim_{u \to 0} \left[ \frac{1-u \cot(u)}{u}\right]\right)$$
By L'Hôpital's rule again, this is $$\exp \left( \lim_{u \to 0} \left[ - \cot(u) + u \csc^2(u) \right]\right)$$
which is $$\exp \left( -\frac{1}{2}\lim_{u\to 0} \frac{ \sin (2 u)-2 u}{\sin ^2(u)} \right)$$
Apply L'Hôpital's rule twice more and obtain
$$ \exp \left( -\frac{1}{2} \lim_{u \to 0} \left[ \frac{4 \sin(2u)}{2 \cos^2 u - 2 \sin^2 u}\right]\right)$$
Finally the limit is not of an indeterminate form, and the limit expression comes to 0. Therefore the original limit is 1.
