Find $\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$ 
Find $$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$$

$$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)
= \lim\limits_{x \rightarrow \infty}  \frac{e^{-\frac{1}{x}}-1}{x^{-0.5}}
= \lim\limits_{x \rightarrow \infty}  \frac{[e^{-x^{-1}}-1]'}{[x^{-0.5}]'}
= \lim\limits_{x \rightarrow \infty}  \frac{x^{-2}e^{-x^{-1}}}{-0.5x^{-1.5}}
= 2 \cdot \lim\limits_{x \rightarrow \infty} \frac{1}{e^{x^-1}x^{0.5}}
$$ 
So as $x \rightarrow \infty$, $\frac{1}{e^{x^-1}x^{0.5}} \rightarrow \frac{1}{1 \cdot \infty} \rightarrow 0$
Is this correct ? any input is much appreciated

Note the fundamental limit formula $$\lim_{t\to 0}\frac{a^{t}-1}{t}=\log a$$ and put $x=1/t,a=1/e$ to get $$\lim_{x\to \infty} x(e^{-1/x}-1)=-1$$ and then dividing by $\sqrt{x} $ we can see that the desired limit is $0$.
