Evaluating the $\lim_{x \to \infty}(e^x + x)^{\frac{1}{x}}$ as $x$ approaches $\infty$ I am trying to evaluate $\lim_{x \to \infty}(e^x + x)^{\frac{1}{x}}$. So, I set $t = \cfrac{1}{x}.$ We know that $x \to \infty \iff \cfrac{1}{x} \to 0^{+}$.
Then, we can say that $\lim_{x \to \infty}(e^x + x)^{\frac{1}{x}}= \lim_{t \to 0^{+}}(e^{\frac{1}{t}}+t)^{t}$. Now I am stuck. I can not find any functions to use Squeeze theorem, so what can I do after this point?
 A: $$\left(e^x+x\right)^{1/x}=e\left[\left(1+\frac1{e^x/x}\right)^{e^x/x}\right]^{1/e^x}\xrightarrow[x\to\infty]{}e\cdot e^0=e$$
A: Elementary squeeze theorem!
$$e=(e^x+0)^{1/x}<(e^x+x)^{1/x}<(e^x+e^x)^{1/x}=e\sqrt[x]2$$
And it is fairly easy to show that $\lim_{x\to\infty}\sqrt[x]2=1$, hence

$$\lim_{x\to\infty}(e^x+x)^{1/x}=e$$

A: $e^{-x}\leq1$ then $1+xe^{-x}\leq1+x$ and $(1+xe^{-x})^\frac1x\leq(1+x)^\frac1x$ so from $\lim_{x\to\infty}\dfrac{\ln(x+1)}{x}\to0$ we have
$$\lim_{x\to\infty}e(1+xe^{-x})^\frac1x\leq \lim_{x\to\infty}e(1+x)^\frac1x\leq e$$
on the other hand with Bernoulli inequality
$$\lim_{x\to\infty}e(1+xe^{-x})^\frac1x\geq \lim_{x\to\infty}e(1+\frac1xxe^{-x})\to e$$
A: Compute the limit of the log: $\;\dfrac1x\ln(\mathrm e^x+x)$, and use  equivalents: since $x=o(\mathrm e^x)$,
$$\mathrm e^x+x\sim_\infty\mathrm e^x,\enspace\text{hence}\quad\ln(\mathrm e^x+x)\sim_\infty\ln(\mathrm e^x)=x, \enspace\text{so}\dfrac1x\ln(\mathrm e^x+x)\sim_\infty1.$$
A: Observe
$$
\frac{\log(e^x+x)}{x} = \frac{\log(1+\frac{x}{e^x})+x}{x}
$$
We know that 
$$
\lim_{x\rightarrow\infty}\log(1+\frac{x}{e^x}) = \log1=0
$$
Therefore 
$$
\lim_{x\rightarrow\infty}\frac{\log(1+\frac{x}{e^x})+x}{x} = 1
$$
Finally, the original limit
$$
\lim_{x\rightarrow\infty}(e^x+x)^{\frac{1}{x}} = \exp{\left\{\lim_{x\rightarrow\infty}\frac{\log(e^x+x)}{x}\right\}} = e
$$
this is a way to look at it without L'Hopital or many other advanced limit techniques. 
