What's $\lim_{x \to 0} 2^{\frac{1}{x}}\frac{x-1}{x-2}$? I need to find the following limit
\begin{align}
\lim_{x \to 0} \left\{f(x) = 2^{\frac{1}{x}}\frac{x-1}{x-2}\right\}
\end{align}
After plotting this thing I got $\lim_{x\to 0^-} f(x) = 0$ and $\lim_{x\to 0^+} f(x) = \infty$.
I am not too sure, however, how one would show it analytically. 
 A: The factor $(x-1)/(x-2)$ is well behaved around zero, the problem is $2^{1/x}$, note that
$$
2^{1/x} = e^{\ln 2^{1/x}} = e^{\ln 2/x}
$$
and remember that 
$$
\lim_{x\to +\infty} e^x = +\infty ~~~~\mbox{and}~~~ \lim_{x\to -\infty} e^x = 0
$$
With these two things together
$$
\lim _{x\to 0^+}2^{1/x} = \lim _{x\to 0^+} e^{\ln2 / x} = \lim _{y\to +\infty} e^{y\ln 2 } = +\infty
$$
whereas
$$
\lim _{x\to 0^-}2^{1/x} = \lim _{x\to 0^-} e^{\ln2 / x} = \lim _{y\to -\infty} e^{y\ln 2 } = 0
$$
A: Change variable as $x\rightarrow\frac{1}{x}$ and you will get
$$
\lim_{x\rightarrow\infty}2^x\frac{1-x}{1-2x}
$$
and you should see the result.
A: $$\lim_{x \to 0}2^{\frac{1}{x}}\times \underbrace{\left( \frac{x-1}{x-2}\right)}_{\text{No indeterminacy here}}=\lim_{x \to 0}\frac{1}{2}2^{\frac{1}{x}} $$
Now put for $x \to 0^+$ i.e. $x=0+h$
$$\lim_{x \to 0^+}\frac{1}{2}2^{\frac{1}{h}} \longrightarrow 2^{\infty} \longrightarrow \infty$$
Now put for $x \to 0^-$ ,$x=0-h$
$$\lim_{x \to 0^-}\frac{1}{2}2^{\frac{1}{-h}}\longrightarrow 2^{-\infty} \longrightarrow 0$$
A: You can split the calculation:
$$\lim_{x\to 0^+}2^{1/x}=\infty$$
$$\lim_{x\to 0^-}2^{1/x}=0$$
so the above limit doesn't exist.
Also,
$$\lim_{x\to 0}\frac{x-1}{x-2}=\frac{1}{2}$$
and then (using that $\lim_{x\to x_0}f(x)\cdot g(x)=\lim_{x\to x_0}f(x)\cdot \lim_{x\to x_0}g(x)$), we get:
$$\lim_{x\to 0^+}2^{1/x}\frac{x-1}{x-2}=\infty$$
$$\lim_{x\to 0^-}2^{1/x}\frac{x-1}{x-2}=0$$
so the limit doesn't exist.
