Question

find the indicated limit. note that lhopitals rule does not apply to every problem, and some problems will require more than one application of lhopitals rule. use -∞ or ∞ when appropriate.
lim(x→∞) (2x² / e^(4x))
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. lim(x→∞) (2x² / e^(4x)) = (type an exact answer in simplified form.)
b. the limit does not exist.

Explanation:

Step1: Check form of limit

As $x\to\infty$, we have $\frac{2x^{2}}{e^{4x}}$ in $\frac{\infty}{\infty}$ form, so L'Hopital's rule can be applied.

Step2: Apply L'Hopital's rule once

Differentiate numerator and denominator. $\frac{d}{dx}(2x^{2}) = 4x$ and $\frac{d}{dx}(e^{4x})=4e^{4x}$. So the limit becomes $\lim_{x\to\infty}\frac{4x}{4e^{4x}}=\lim_{x\to\infty}\frac{x}{e^{4x}}$, still in $\frac{\infty}{\infty}$ form.

Step3: Apply L'Hopital's rule again

Differentiate numerator and denominator again. $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(e^{4x})=4e^{4x}$. So the limit becomes $\lim_{x\to\infty}\frac{1}{4e^{4x}}$.

Step4: Evaluate the limit

As $x\to\infty$, $e^{4x}\to\infty$, so $\lim_{x\to\infty}\frac{1}{4e^{4x}} = 0$.

Answer:

A. $\lim_{x\to\infty}\frac{2x^{2}}{e^{4x}}=0$