Evaluating the $\lim_{x \to \infty}(e^x + x)^{\frac{1}{x}}$ as $x$ approaches $\infty$

Question

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?

Answer 1

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

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

Answer 2

$e^{-x}\leq1$ 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 $$\lim_{x\to\infty}e(1+xe^{-x})^\frac1x\geq \lim_{x\to\infty}e(1+\frac1xxe^{-x})\to e$$

Answer 3

$$\left(e^x+x\right)^{1/x}=e\left[\left(1+\frac1{e^x/x}\right)^{e^x/x}\right]^{1/e^x}\xrightarrow[x\to\infty]{}e\cdot e^0=e$$

Answer 4

Compute the limit of the log: $\;\dfrac1x\ln(\mathrm e^x+x)$, and use equivalents: since $x=o(\mathrm e^x)$, $$\mathrm e^x+x\sim_\infty\mathrm e^x,\enspace\text{hence}\quad\ln(\mathrm e^x+x)\sim_\infty\ln(\mathrm e^x)=x, \enspace\text{so}\dfrac1x\ln(\mathrm e^x+x)\sim_\infty1.$$

Answer 5

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.