Find $\lim_{x \to 0} (\sin x)^{1/x} + (1/x)^{\sin x}$ Evaluate the limit $\lim_{x \to 0} (\sin x)^{1/x} + (1/x)^{\sin x}$
This limit is of the form $(0)^{\infty} + {\infty}^{0}$
I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.
But if first term is an indeterminate form then how would i calculate the limit?
 A: $$
L = \lim_{x\to 0^+} \left((\sin x)^{1\over x} + \left({1\over x}\right)^{\sin x}\right) \\
= \lim_{x\to 0^+} \frac{(\sin x)^{1\over x}\cdot x^{\sin x} + 1}{x^{\sin x}}
$$
By $\sin x \sim x$ as $x\to 0$:
$$
\frac{(\sin x)^{1\over x}\cdot x^{\sin x} + 1}{x^{\sin x}} \sim \frac{x^{1\over x}x^x + 1}{x^x} = \frac{x^{x+{1\over x}}+ 1}{x^x}
$$
By the fact that $\lim_{x\to 0^+} x^x = 1$:
$$
\begin{align}
\lim_{x\to 0^+} \frac{x^{x+{1\over x}}+ 1}{x^x} &= \lim_{x\to 0^+} \left(x^{x+{1\over x}} + 1\right) \\
&= 1+ \lim_{x\to 0^+} x^x \cdot x^{1\over x} \\
&= 1 + \lim_{x\to 0^+}x^x \cdot \lim_{x\to 0^+} \sqrt[x]{x}\\
&= 1 + 1\cdot 0 = 1
\end{align}
$$
A: I'm assuming $x\rightarrow 0^+$ rather than $x\rightarrow\infty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $a\in ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=\sin(x)$:
$$0<\sin(x)^{1/x}\leq\sin(x)$$
By the squeeze theorem we obtain 
$$\lim_{x\rightarrow 0^+}\sin(x)^{1/x}=0$$
