Question

question 4 · 1 point evaluate the following limit using lhospitals rule. enter an exact answer. provide your answer below. $lim_{x
ightarrowinfty}\frac{2x^{2}-14x + 8}{7e^{13x}+9x}$

Explanation:

Step1: Check form of limit

As $x\to\infty$, the limit $\lim_{x\to\infty}\frac{2x^{2}-14x + 8}{7e^{13x}+9x}$ is in the $\frac{\infty}{\infty}$ form.

Step2: Apply L'Hospital's rule

Differentiate the numerator and denominator. The derivative of the numerator $y_1=2x^{2}-14x + 8$ is $y_1'=4x-14$, and the derivative of the denominator $y_2=7e^{13x}+9x$ is $y_2'=91e^{13x}+9$. So the limit becomes $\lim_{x\to\infty}\frac{4x - 14}{91e^{13x}+9}$.

Step3: Apply L'Hospital's rule again

Since it is still in the $\frac{\infty}{\infty}$ form. The derivative of the new - numerator $y_1'=4x-14$ is $y_1'' = 4$, and the derivative of the new - denominator $y_2'=91e^{13x}+9$ is $y_2''=1183e^{13x}$. So the limit becomes $\lim_{x\to\infty}\frac{4}{1183e^{13x}}$.

Step4: Evaluate the limit

As $x\to\infty$, $e^{13x}\to\infty$, so $\lim_{x\to\infty}\frac{4}{1183e^{13x}} = 0$.

Answer:

$0$