verifying $\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x$

There is another way to compute your limit:

$\lim\limits_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^x=$

$=\lim\limits_{x \to +\infty} \left[\left(1+\frac{1}{\sqrt{x}}\right)^\sqrt{x}\right]^\sqrt{x}=$

$=\left[\lim\limits_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^\sqrt{x}\right]^{\lim\limits_{x \to +\infty} \sqrt{x}}=$

$=e^{+\infty}=$

$=+\infty$.

You can show the limit much quicker by bounding your expression from below using the Bernoulli inequality :

$$\left(1+\frac{1}{\sqrt{x}}\right)^x\geq 1+x\cdot\frac{1}{\sqrt{x}} =1+\sqrt x\stackrel{x\to +\infty}{\longrightarrow}+\infty$$

I think you ought to specify $\lim_{t\to 0^+}$, just to be on the safe side. Apart from that, it looks correct.

It's a lot quicker, however, to use $s^2=x$ and note that for any real number $k$, we have $$ \lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^n =\lim_{s\to\infty}\left(1+\frac1{s}\right)^{s^2}\\ \geq \lim_{s\to\infty}\left(1+\frac1{s}\right)^{ks}=e^k $$