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.

$\underset{x\to \infty }{\lim }\left(e^x-\frac{e^x}{x+1}\right)$

Short answer

The value of the limit is$\infty$

Step-by-step solution

Step 1. Given information

Given function$\underset{x\to \infty }{\lim }\left(e^x-\frac{e^x}{x+1}\right)$

Apply L-Hospital rule and solve

Calculating, we get

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