Question

Show that each of the following limits is of the form $\frac{0}{0}$and then use L’Hopital’s rule to calculate the limit: ˆ

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

Short answer

The solution is$\frac{3\log \left(3\right)}{-2}$

Step-by-step solution

Step 1. Given information

The given limit is$\underset{x\to 1}{\lim }\frac{3^x-3}{1-x^2}$

Step2. Calculate limit

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{3^x-3}{1-x^2} \\ \underset{x\to 1}{\lim }\frac{{\displaystyle \frac{d\left(3^x-3\right)}{dx}}}{{\displaystyle \frac{d\left(1-x^2\right)}{dx}}}=\underset{x\to 1}{\lim }\frac{3^x\log \left(3\right)}{-2x} \\ =\frac{3\log \left(3\right)}{-2} \end{gathered}$$

Step 3. The solution 

The solution is$\frac{3\log \left(3\right)}{-2}$