Question: Evaluate $\lim\_{x\to \infty}(1+\frac{1}{\sqrt{x}})^x$

This is an exercise in my analysis book and I'm having trouble solving it. It's clearly related to the limit of $e$. I've tried rewriting:

Evaluate $\lim\_{x\to \infty}((1+\frac{1}{\sqrt{x}})^\sqrt{x})^\sqrt{x}$.

Now I sadly cannot rewrite this as $\lim\_{x\to \infty}(e)^x$ or anything similar. Thoughts?

Answer: 
Note that 

$$\lim\_{x\to \infty}\left(1+\frac1{\sqrt{x}}\right)^{\sqrt{x}}=e $$

Therefore, for sufficiently large $x$, $\left(1+\frac1{\sqrt{x}}\right)^{\sqrt{x}}\ge 2$ (i.e., take $\epsilon = e-2>0$). 

Hence, for sufficiently large $x$, we have

$$\left(1+\frac1{\sqrt{x}}\right)^{x}\ge 2^{\sqrt{x}}\to \infty\,\,\text{as}\,\,x\to \infty$$

And we are done!

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Alternatively, in [THIS ANSWER](https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263), I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm satisfies the inequality 

$$\log(x)\ge \frac{x}{1+x}$$

Therefore, 

$$\left(1+\frac1{\sqrt{x}}\right)^{x}\ge e^{\frac{x}{1+\sqrt{x}}}\to \infty\,\,\text{as}\,\,x\to \infty$$
For $\alpha\ge1, t\ge-1$, $(1+t)^\alpha\ge1+\alpha t$ 

We can see this by considering the tangent at the origin and convexity. (This is Bernoulli's inequality.) 

Take $t=\frac{1}{\sqrt{x}},\alpha=x$ to see $(1+\frac{1}{\sqrt{x}})^x\ge1+\sqrt{x}\to\infty$.