evaluate the limit. lim x→∞ e5x / x3

Explanation:

Step1: Identify the form

As $x\to\infty$, we have the indeterminate - form $\frac{\infty}{\infty}$ since $\lim_{x\to\infty}e^{5x}=\infty$ and $\lim_{x\to\infty}x^{3}=\infty$. So, we can apply L'Hopital's rule.

Step2: Apply L'Hopital's rule once

Differentiate the numerator and denominator. The derivative of $y = e^{5x}$ is $y^\prime=5e^{5x}$ (by the chain - rule, if $y = e^{u}$ and $u = 5x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=e^{u}\cdot5 = 5e^{5x}$), and the derivative of $y=x^{3}$ is $y^\prime = 3x^{2}$. So, $\lim_{x\to\infty}\frac{e^{5x}}{x^{3}}=\lim_{x\to\infty}\frac{5e^{5x}}{3x^{2}}$.

Step3: Apply L'Hopital's rule again

Since $\lim_{x\to\infty}\frac{5e^{5x}}{3x^{2}}$ is still in the $\frac{\infty}{\infty}$ form, differentiate the numerator and denominator again. The derivative of $5e^{5x}$ is $25e^{5x}$ and the derivative of $3x^{2}$ is $6x$. So, $\lim_{x\to\infty}\frac{5e^{5x}}{3x^{2}}=\lim_{x\to\infty}\frac{25e^{5x}}{6x}$.

Step4: Apply L'Hopital's rule a third time

Since $\lim_{x\to\infty}\frac{25e^{5x}}{6x}$ is in the $\frac{\infty}{\infty}$ form, differentiate the numerator and denominator again. The derivative of $25e^{5x}$ is $125e^{5x}$ and the derivative of $6x$ is $6$. So, $\lim_{x\to\infty}\frac{25e^{5x}}{6x}=\lim_{x\to\infty}\frac{125e^{5x}}{6}$.

Step5: Evaluate the limit

As $x\to\infty$, $e^{5x}\to\infty$. So, $\lim_{x\to\infty}\frac{125e^{5x}}{6}=\infty$.

Answer:

$\infty$