Q. 33 - 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.

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

Step-by-Step Solution

Answer

The value of limit is zero

Step 1. Given information

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

Step 2: Apply L-Hospital rule and solve

Calculating, we get

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