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
 A: Yes your way is correct,but you can do it simply as  $$\lim _{ x\rightarrow \infty  } \sqrt { x } \left( e^{ -\frac { 1 }{ x }  }-1 \right) =\lim _{ x\rightarrow \infty  } \sqrt { x } \frac { \left( e^{ -\frac { 1 }{ x }  }-1 \right)  }{ -\frac { 1 }{ x }  } \left( -\frac { 1 }{ x }  \right) =\lim _{ x\rightarrow \infty  }{ -\frac { \sqrt { x }  }{ x }  } =0$$
A: 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 $0$.
A: I always prefer Taylor expansion to L'Hospital rule, as it is clearer and more generalized. For this one, write:
\begin{align}
\sqrt{x}(e^{-1/x} - 1) & = \sqrt{x}\left(1 - \frac{1}{x} + o\left(\frac{1}{x}\right) - 1\right) = -\frac{1}{\sqrt{x}} + o\left(\frac{1} 
{\sqrt{x}}\right) \\
& \to 0 \quad \text{as } x \to \infty.
\end{align}
It might be safe to say that all univariate limit problems/exercises can be handled neatly by applying Taylor's theorem with the Maclaurin remainder. 
