Question

18 mark for review
$lim_{x
ightarrowinfty}\frac{x^{3}}{e^{2x}}$ is
a 0
b $\frac{2}{9}$
c $\frac{2}{3}$
d 1

Explanation:

Step1: Apply L'Hopital's rule

Since $\lim_{x
ightarrow\infty}\frac{x^{3}}{e^{2x}}$ is in the $\frac{\infty}{\infty}$ form, by L'Hopital's rule, we differentiate the numerator and denominator. The derivative of $x^{3}$ is $3x^{2}$ and the derivative of $e^{2x}$ is $2e^{2x}$, so we get $\lim_{x
ightarrow\infty}\frac{3x^{2}}{2e^{2x}}$.

Step2: Apply L'Hopital's rule again

This is still in the $\frac{\infty}{\infty}$ form. Differentiating again, the derivative of $3x^{2}$ is $6x$ and the derivative of $2e^{2x}$ is $4e^{2x}$, so we have $\lim_{x
ightarrow\infty}\frac{6x}{4e^{2x}}$.

Step3: Apply L'Hopital's rule one - more time

It's still $\frac{\infty}{\infty}$. Differentiating, the derivative of $6x$ is $6$ and the derivative of $4e^{2x}$ is $8e^{2x}$, so we get $\lim_{x
ightarrow\infty}\frac{6}{8e^{2x}}$.

Step4: Evaluate the limit

As $x
ightarrow\infty$, $e^{2x}
ightarrow\infty$, so $\lim_{x
ightarrow\infty}\frac{6}{8e^{2x}} = 0$.

Answer:

A. 0