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: 
Suppose $x\in [n,n+1),$ where $n$ is a positive integer. Using the binomial theorem we have 

$$(1+1/\sqrt x)^x > (1+1/\sqrt {n+1})^n = 1^n + n\cdot 1^{n-1}\cdot (1/\sqrt {n+1})^1 + \cdots > n/\sqrt {n+1}.$$

Since $n/\sqrt {n+1} \to \infty,$ the desired limit is $\infty.$
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, 

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