Limit of the given expression. For $x>0$,$$\lim_{x\to 0}((\sin(x))^{\frac{1}{x}}+(\frac{1}{x})^{\sin(x)})$$ is?. So now I calculated limits individually. Let $\lim_{x\to 0} ((\sin(x))^{1/x})=y$ thus I took log to get $\frac{1}{x}(1-(1-\sin(x))....$=1 (after writing $\sin(x)=1-(1-\sin(x))$ and then using series expansion for $\ln(1-x)$. So I got it as $1$ thus $y=e$ and similarly for next I got limit value as $1$ thus the value of limit is $e+1$ but correct answer given is $1$ . Thanks
 A: Hint. One may observe that, as $x \to 0^+$,
$$
-\sin x\cdot \log x= -\frac{\sin x}x \cdot x\log x\to -1 \cdot 0= 0
$$ giving

$$
\left(\frac1x \right)^{\large \sin x}=e^{-\sin x\cdot \ln x} \to e^0=1
$$ 

and, as $x \to 0^+$,
$$
\frac{\log(\sin x)}x =\frac1x \cdot\log(\sin x)\to \infty \times (-\infty) =-\infty
$$ giving

$$
\left(\sin x \right)^{\large \frac1x}=e^{\large \frac{\log(\sin x)}x } \to e^{-\infty}=0
$$ 

Thus the sought limit is equal to $1$.
A: Consider $$A=\sqrt[x]{\sin (x)}\implies \log(A)=\frac 1x \log (\sin (x))$$ Now, using Taylor expansion $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)=x\left(1-\frac{x^2}{6}+O\left(x^4\right)\right)$$ $$\log (\sin (x))=\log (x)+\log\left(1-\frac{x^2}{6}+O\left(x^4\right)\right)=\log (x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(A)=\frac 1x \log (\sin (x))=\frac{\log (x)}{x}-\frac{x}{6}+O\left(x^3\right)$$ So,$\log(A)\to -\infty$ that is to say $A\to 0$
A: This is implicitly a one-sided limit ($x\to0^+$). Consider the first piece:
$$
\lim_{x\to0}(\sin x)^{1/x}
$$
and compute instead the limit of the logarithm:
$$
\lim_{x\to0}\frac{\log\sin x}{x}=-\infty
$$
gives $\lim_{x\to0}(\sin x)^{1/x}=0$.
For the second summand you can consider $\lim_{x\to0}x^{\sin x}$ and take the logarithm:
$$
\lim_{x\to0}\sin x\log x=
\lim_{x\to0}\frac{\sin x}{x}(x\log x)=1\cdot0=0
$$
Thus we get $\lim_{x\to0}x^{\sin x}=1$ and finally
$$
\lim_{x\to 0}
\left(
(\sin x)^{1/x}+\left(\frac{1}{x}\right)^{\sin x}\right)=0+\frac{1}{1}=1
$$
