Evaluating $\lim_{x \to 0+} \left[ \sin(x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right] $? For $x>0$, $$\lim_{x \rightarrow 0} \left[ \sin(x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right]
$$These are two forms of $0^\infty$ and $\infty^0$. I know these are to be evaluated separately and then added. But how do I start?
 A: At first start with 
$$a^b=e^{b\cdot \ln(a)}$$ 
(this is the definition of $a^b$ for $a>0$).
and because of the $e$ function is continuous 
$$\lim e^{x_n}=e^{\lim x_n}$$
This is because of the sequence definition or test for continuous functions, a function $f$ is continuous if for every convergent sequence $x_n$ 
$$ \lim_{n\rightarrow \infty} f(x_n)=f(\lim_{n\rightarrow \infty} x_n)$$
Since both term are greater equal we can check the first, and afterwards the second. 
$$\sin(x)^\frac{1}{x}=e^{\ln(\sin(x))\cdot \frac{1}{x}}$$
Because of it is continuous we check
$$\lim_{x\rightarrow 0}\frac{\ln(\sin(x))}{x}$$
As we have a $\frac{-\infty}{\infty}$ expression here, we can use L'hospital 
$$\lim_{x\rightarrow 0 } \frac{\ln(\sin(x))}{x}=-\lim_{x\rightarrow 0} \frac{\cos(x)}{\sin(x)}=-\infty$$ (the $-$ must be there because $\ln(\sin(x))<0$ as $x\rightarrow 0$
And because $$\lim_{x\rightarrow - \infty} e^{x}=0$$ the first term vanishes.
The second Term is handled like this:
$$\left(\frac{1}{x}\right)^{\sin(x)} = e^{\ln\left(\frac{1}{x}\right) \sin(x)}$$
Now 
$$\ln\left(\frac{1}{x}\right) \cdot \sin(x)=\frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{\sin(x)}}$$
with L'hospital 
$$\lim_{x\rightarrow 0} \frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{\sin(x)}}=
\lim_{x\rightarrow 0} \frac{x}{\cos(x)}=0 $$
so the second term is $e^0$ so the limit is 1.
A: Note that $0^{\infty}$ is not an indeterminate form. Then
$$\lim_{x \rightarrow 0^+} \left[ \sin(x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right]=\lim_{x \rightarrow 0^+} \left(\frac{1}{x}\right)^{\sin(x)}=\lim_{y \rightarrow +\infty}y^{\sin(1/y)}=\lim_{y \rightarrow +\infty}(y^{1/y})^{y\sin(1/y)}=1$$
The other limit to $0^-$is complex infinity.
