Q. 26 - 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 

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

Step-by-Step Solution

Answer

The value of the given limit $\underset{x\to \infty }{\lim }\frac{x+3^x}{3-2^x}=-\infty$

Step 1. Given information

The given limit $\underset{x\to \infty }{\lim }\frac{x+3^x}{3-2^x}$

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

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x+3^x}{3-2^x}=\underset{x\to \infty }{\lim }\frac{1+3^x \ln 3}{0-2^x\ln 2} \\ =\underset{x\to \infty }{\lim }\frac{1+3^x \ln 3}{-2^x \ln 2} \\ =-{\left(\frac{\ln 3}{\ln 2}\right)}^2\underset{x\to \infty }{\lim }\frac{3^x}{2^x} \\ =-{\left(\frac{\ln 3}{\ln 2}\right)}^2\underset{x\to \infty }{\lim }{\left(\frac{3}{2}\right)}^x \\ =-{\left(\frac{\ln 3}{\ln 2}\right)}^2\underset{x\to \infty }{\lim }{\left(1.5\right)}^x \\ =-\infty \end{gathered}$$

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