Evaluate $\lim_{x\to\infty}\left (\frac{1}{x}+2^{\frac{1}{x}}\right)^x$ $$
\Large\lim_{x\to\infty}\left(\frac{1}{x}+2^{\frac{1}{x}}\right)^x\Large\\
=\Large\lim_{x\to\infty}\left(1+\left[\frac{1}{x}+2^{\frac{1}{x}}-1\right]\right)^x\Large\\
=\Large\lim_{x\to\infty}\left(1+\left[\frac{1}{x}+2^{\frac{1}{x}}-1\right]\right)^{\frac{1}{\frac{1}{x}+2^{\frac{1}{x}}-1}\color{red}{\left(1+x2^{\frac{1}{x}}-x\right)}}\Large\\
=\Large\lim_{x\to\infty} e^{\color{red}{\left(1+x2^{\frac{1}{x}}-x\right)}}
$$
I got trouble getting further (especially dealing with the type $0\cdot\infty$) and how to get around with it, please help, thank you for any comments.
 A: Taking the logarithm of the function and setting $t:=\dfrac1x$, we can use L'Hospital:
$$\lim_{t\to0}\frac{\log(t+2^t)}{t}=\lim_{t\to0}\frac{(1+2^t\log2)}{t+2^t}.$$

$=1+\log 2$, giving the final answer $2e$.

A: You have that
$$e^{1+x 2^{\frac{1}{x}}-x}=e^{1+x\left(2^{\frac{1}{x}}-1\right)}$$
Now recall that
$$\lim_{t \to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{e^{t \log a}-1}{t}=\lim_{t \to 0} \frac{e^{t \log a}-1}{t \log a} \cdot \log a$$
By putting $u=t \log a$ and noticing that $t \log a \to 0$ as $t \to 0$, hence
$$\lim_{t \to 0} \frac{e^{t \log a}-1}{t \log a} \cdot \log a=\lim_{u \to 0} \frac{e^u-1}{u} \log a =1 \cdot \log a =\log a$$
So, in $e^{1+x\left(2^{\frac{1}{x}}-1\right)}$, putting $z=\frac{1}{x}$ you have that $z \to 0$ as $x \to \infty$ and so
$$\lim_{x \to \infty} e^{1+x\left(2^{\frac{1}{x}}-1\right)}=\lim_{z \to 0^+} e^{1+\frac{2^z-1}{z}}=e^{1+\log 2}=e \cdot e^{\log 2}=2e$$
Alternatively, notice that since when $x \to \infty$ it is $\frac{1}{x}+2^{1/x}>0$, it is
$$\left(\frac{1}{x}+2^{1/x}\right)^x=\exp\left[x \log \left(\frac{1}{x}+2^{1/x}\right)\right]=\exp \left \{x \left[\frac{1}{x}\log 2 + \log\left(1+\frac{1}{x\cdot2^{1/x}}\right)\right]\right\}$$
Now use that $\log(1+t)=t+\text{o}(t)$ as $t \to 0$; I'm using $\exp \left[f(x)\right]=e^{f(x)}$ for better formatting.
A: As $x\to\infty$ we have $1/x+2^{1/x}=$ $1/x+e^{(\ln 2)/x}=$ $1/x+1+(\ln 2)/x+O(1/x^2).$
So $\lim_{n\to\infty} (1/x+2^{1/x})^{\,x}=$ $\lim_{n\to\infty}(\,1+\frac {1+\ln 2}{x}+O(1/x^2)\,)^{\,x}=$
$=\lim_{n\to\infty}(1+\frac {1+\ln 2}{x})^x=$ $e^{1+\ln 2}=2e.$
