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.

$\underset{x\to \infty }{\lim }\frac{x{e}^{-3x}}{x^2+3x+1}$

Short answer

The value of limit is$0$

Step-by-step solution

Step 1. Given information

Given function$\underset{x\to \infty }{\lim }\frac{x{e}^{-3x}}{x^2+3x+1}$

Apply L-Hospital rule and solve

Calculating, we get

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x{e}^{-3x}}{x^2+3x+1}=\underset{x\to \infty }{\lim }\frac{x}{e^{3x}\left(x^2+3x+1\right)} \\ \underset{x\to \infty }{\lim }\frac{x{e}^{-3x}}{x^2+3x+1}=\underset{x\to \infty }{\lim }\frac{x}{e^{3x}\left(x^2+3x+1\right)} \\ =\underset{x\to \infty }{\lim }\frac{1}{e^{3x}(2x+3)+\left(x^2+3x+1\right)·e^{3x}·3} \\ =\underset{x\to \infty }{\lim }\frac{1}{e^{3x}\left[2x+3+3x^2+9x+3\right]} \\ =\underset{x\to \infty }{\lim }\frac{1}{e^{3x}\left(3x^2+11x+6\right)} \\ =0 \end{gathered}$$