Q. 24 - Calculus

Question

Calculate each of the limits in Exercises 21–48. Some of these limits are made easier by L’Hopital’s rule, and some are not 

$\underset{x\to \infty }{\lim }\frac{e^x-7}{8x^2+12x+5}$

Step-by-Step Solution

Answer

The value of the given limit $\underset{x\to \infty }{\lim }\frac{e^x-7}{8x^2+12x+5}=\infty$

Step 1. Given information

The given limit$\underset{x\to \infty }{\lim }\frac{e^x-7}{8x^2+12x+5}$

Step 2. Using L’Hopital’s rule find the value of given limit

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{e^x-7}{8x^2+12x+5}=\underset{x\to \infty }{\lim }\frac{e^x-0}{8.2x+12.1+0} \\ =\underset{x\to \infty }{\lim }\frac{e^x}{16x+12} \\ =\underset{x\to \infty }{\lim }\frac{e^x}{16.1+0} \\ =\underset{x\to \infty }{\lim }\frac{e^x}{12} \\ =\frac{e^{\infty }}{16} \\ =\infty \end{gathered}$$

Therefore,the value of the given limit $=\infty$