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 $\infty.$
Compute first the limit of the (natural) logarithm:
$$
\lim\_{x\to\infty}\log\left(\left(1+\frac{1}{\sqrt{x}}\right)^{\!x}\right)=
\lim\_{x\to\infty}x\log\left(1+\frac{1}{\sqrt{x}}\right)=
\lim\_{t\to0^+}\frac{\log(1+t)}{t^2}
$$
after the substitution $\sqrt{x}=1/t$, with $t>0$. Now
$$
\lim\_{t\to0^+}\frac{\log(1+t)}{t}=1
$$
so
$$
\lim\_{t\to0^+}\frac{\log(1+t)}{t^2}=\infty
$$
Therefore also your limit is $\infty$, because $\lim\_{z\to\infty}e^z=\infty$.