Q. 35 - 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.

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

Step-by-Step Solution

Answer

The value of limit is $0$

Step 1. Given information

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

Step 2: Apply L-Hospital rule and solve

Calculating, we get

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