Answer
$\infty$
Work Step by Step
Given: $\lim\limits_{u\to\infty}\dfrac{e^{u/10}}{u^3}$
Here, $u\to\infty;e^{u/10}\to\infty;u^3\to\infty$.
This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will apply L'Hospital Rule.
or, $=\lim_{u\to\infty}\dfrac{(e^{u/10})'}{(u^3)'}$
or, $=\lim\limits_{u\to\infty}\dfrac{e^{u/10}(u/10)'}{3u^2}$
or,$A=\lim\limits_{u\to\infty}\dfrac{e^{u/10}}{30u^2}$
Here,$u\to\infty;e^{u/10}\to\infty; u^2\to\infty$.
Again, This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will reapply L'Hospital Rule.
$=\lim\limits_{u\to\infty}\dfrac{e^{u/10}(u/10)'}{60u}$
and $=\lim\limits_{u\to\infty}\dfrac{e^{u/10}}{600u}$
Again, as $u\to\infty$, $e^{u/10}\to\infty$ and $u\to\infty$.
Again, this shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will reapply L'Hospital Rule.
Thus we have
$=\lim_{u\to\infty}\dfrac{e^{u/10}(u/10)'}{600}$
and $=\lim\limits_{u\to\infty}\dfrac{e^{(u/10)}}{6000}$
As $u\to\infty$; $e^{u/10}\to\infty;\dfrac{e^{(u/10)}}{6000} \to \infty.$
Hence, we have
$=\infty$
