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: 
$y= (1+\frac{1}{\sqrt{x}})^x$ then $\ln y=\ln(1+\frac{1}{\sqrt{x}})^x=x\ln(1+\frac{1}{\sqrt{x}})$.
We have 
$$\lim\_{x\to\infty} x\ln(1+\frac{1}{\sqrt{x}})=\lim\_{x\to\infty}=\frac{\ln (1+\frac{1}{\sqrt{x}})}{\frac{1}{x}}=^H\lim\_{x\to\infty}\frac{ \frac{\frac{-1}{2x\sqrt{x}}}{\frac{\sqrt{x}+1}{\sqrt{x}}}}{\frac{-1}{x^2}}=\lim\_{x\to\infty}\frac{ \frac{-1}{2x(\sqrt{x}+1)}}{\frac{-1}{x^2}}=\infty$$
So $\lim\_{x\to\infty} \ln y=\infty$ and consequently 
$$\lim\_{x\to\infty} y=\infty$$
It is not hard to show that for $x>1$, $$\left(1+\frac1{\sqrt x}\right)^{\sqrt x}>2$$ and your limit is $\infty$.