find the indicated limit. note that lhopitals rule does not apply to every problem, and some problems will…
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.\nlim(x→∞) (2x² / e^(4x))\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. lim(x→∞) (2x² / e^(4x)) = (type an exact answer in simplified form.)\nb. the limit does not exist.
Answer
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. $\lim_{x\to\infty}\frac{2x^{2}}{e^{4x}}=0$