Question

use lhôpitals rule to find the following limit
lim_{x
ightarrowinfty}7x^{2}e^{-x}
lim_{x
ightarrowinfty}7x^{2}e^{-x}=square \text{ (type an exact answer.)}

Explanation:

Step1: Rewrite the function

Rewrite $\lim_{x
ightarrow\infty}7x^{2}e^{-x}$ as $\lim_{x
ightarrow\infty}\frac{7x^{2}}{e^{x}}$. This is in the $\frac{\infty}{\infty}$ - form, so L'Hôpital's rule can be applied.

Step2: Apply L'Hôpital's rule once

Differentiate the numerator and denominator. The derivative of $7x^{2}$ is $14x$ and the derivative of $e^{x}$ is $e^{x}$. So we get $\lim_{x
ightarrow\infty}\frac{14x}{e^{x}}$. This is still in the $\frac{\infty}{\infty}$ - form.

Step3: Apply L'Hôpital's rule again

Differentiate the numerator and denominator again. The derivative of $14x$ is $14$ and the derivative of $e^{x}$ is $e^{x}$. So we have $\lim_{x
ightarrow\infty}\frac{14}{e^{x}}$.

Step4: Evaluate the limit

As $x
ightarrow\infty$, $e^{x}
ightarrow\infty$. So $\lim_{x
ightarrow\infty}\frac{14}{e^{x}} = 0$.

Answer:

$0$