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 
$$\lim\_{x\to\infty} y=\infty$$
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 $(1+\frac{1}{\sqrt{x}})^x\ge1+\sqrt{x}\to\infty$.