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# Limits 2

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Shreya Nair
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Limits 2

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#66079
Saumik Karfa
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take $z=\frac 1x$ Then $z \to \infty$ as $x \to 0$

then the given limit becomes :

$\lim_{z \to \infty} \frac{z}{e^z} [\frac {\infty}{\infty}]$

Now, $\lim_{z \to \infty} \frac{1}{e^z}$ [using L’Hospital]

$=0$

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