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.

$\underset{x\to 2^{+}}{\lim }\ln \left(\frac{2^x-4}{x-2}\right)$.

Short answer

The value of the limit$\underset{x\to 2^{+}}{\lim }\ln \left(\frac{2^x-4}{x-2}\right)$is,$\ln \left(\ln 16\right)$.

Step-by-step solution

Step 1. Given information

$\underset{x\to 2^{+}}{\lim }\ln \left(\frac{2^x-4}{x-2}\right)$.

Step 2. The limit using L'Hopital's rule is given below:

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }\ln \left(\frac{2^x-4}{x-2}\right)=\underset{x\to 2^{+}}{\lim }\ln \left(\frac{2^x-4}{x-2}\right) \\ =\ln \left[\underset{x\to 2^{+}}{\lim }\left(\frac{2^x-4}{x-2}\right)\right] \end{gathered}$$

$=\ln \left[\underset{x\to 2^{+}}{\lim }\left(\frac{2^x\ln 2}{1-0}\right)\right]$[L'Hopital's rule]

$$\begin{gathered} =\ln \left[\underset{x\to 2^{+}}{\lim }\left(2^x\ln 2\right)\right] \\ =\ln \left(2^2\ln 2\right) \\ =\ln \left(4\ln 2\right) \\ =\ln \left(\ln 2^4\right) \\ =\ln \left(\ln 16\right) \end{gathered}$$

The exact value is,$\ln \left(\ln 16\right)$.