<s>[INST] 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
 [/INST] Yes your way is correct,but you can do it simply as  $$\lim _{ x\rightarrow \infty  } \sqrt { x } \left( e^{ -\frac { 1 }{ x }  }-1 \right) =\lim _{ x\rightarrow \infty  } \sqrt { x } \frac { \left( e^{ -\frac { 1 }{ x }  }-1 \right)  }{ -\frac { 1 }{ x }  } \left( -\frac { 1 }{ x }  \right) =\lim _{ x\rightarrow \infty  }{ -\frac { \sqrt { x }  }{ x }  } =0$$
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