Question

question 2. 1 point
what is $lim_{x
ightarrow+infty}p(x)$?
p(x) = $\frac{4x^{2}}{8e^{x}}$
provide your answer below:

Explanation:

Step1: Recall L'Hopital's rule

When $\lim_{x
ightarrow+\infty}\frac{f(x)}{g(x)}$ is in the $\frac{\infty}{\infty}$ form, $\lim_{x
ightarrow+\infty}\frac{f(x)}{g(x)}=\lim_{x
ightarrow+\infty}\frac{f'(x)}{g'(x)}$. Here $f(x) = 8e^{x}$ and $g(x)=4x^{2}$, and $\lim_{x
ightarrow+\infty}8e^{x}=+\infty$, $\lim_{x
ightarrow+\infty}4x^{2}=+\infty$.

Step2: Differentiate numerator and denominator

$f'(x)=8e^{x}$, $g'(x) = 8x$. So $\lim_{x
ightarrow+\infty}\frac{8e^{x}}{4x^{2}}=\lim_{x
ightarrow+\infty}\frac{8e^{x}}{8x}$.

Step3: Apply L'Hopital's rule again

Since $\lim_{x
ightarrow+\infty}8e^{x}=+\infty$ and $\lim_{x
ightarrow+\infty}8x=+\infty$, differentiate again. $f''(x)=8e^{x}$, $g''(x)=8$. So $\lim_{x
ightarrow+\infty}\frac{8e^{x}}{8x}=\lim_{x
ightarrow+\infty}\frac{8e^{x}}{8}$.

Step4: Evaluate the limit

$\lim_{x
ightarrow+\infty}\frac{8e^{x}}{8}=\lim_{x
ightarrow+\infty}e^{x}=+\infty$.

Answer:

$+\infty$