Q. 39 - 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.

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

Step-by-Step Solution

Answer

The exact value of $\underset{x\to 2^{+}}{\lim }\frac{\ln (x-2)}{\ln \left(x^2-4\right)}$ is, $1$.

Step 1 . Given information

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

Step 2 . The limit is given below:

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }\frac{\ln (x-2)}{\ln \left(x^2-4\right)}=\underset{x\to 2^{+}}{\lim }\frac{\ln (x-2)}{\ln \left[\right(x-2\left)\right(x+2\left)\right]} \\ =\underset{x\to 2^{+}}{\lim }\frac{\ln (x-2)}{\ln (x-2)+\ln (x+2)} \\ =\underset{x\to 2^{+}}{\lim }\frac{1}{\frac{\ln (x-2)}{\ln (x+2)}+1} \\ =\frac{1}{0+1} \\ =1 \end{gathered}$$

The exact value is, $1.$