Q. 28 - Calculus

Question

Calculate each of the limits in Exercises 21-48. Some of these limits are made easier by L'Hôpital's rule, and some are not.

$\underset{x\to \infty }{\lim }{x}^2{e}^{-x}$

Step-by-Step Solution

Answer

The value of limit is $\infty$

Step 1. Given information

Given function $\underset{x\to \infty }{\lim }{x}^2{e}^{-x}$

Step 2: Apply L-Hospital rule and solve

Calculating, we get

$$\begin{gathered} \underset{x\to \infty }{\lim }{x}^2{e}^{-x}=\underset{x\to \infty }{\lim }\frac{x^2}{e^x} \\ \underset{x\to \infty }{\lim }{x}^2{e}^{-x}=\underset{x\to \infty }{\lim }\frac{2x}{e^x} \\ \underset{x\to \infty }{\lim }{x}^2{e}^{-x}=\infty \end{gathered}$$