Limits 2

[Shreya Nair]

Limits 2

[Saumik Karfa]

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]

$=0$

[Author]

Posts