Question

Calculate each of the limits in Exercises $49-64$. Some of these limits are made easier by considering the logarithm of the limit first, and some are not.

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

Short answer

The exact value of the limit$\underset{x\to 2^{+}}{\lim }(x-2{)}^{x^2-4}$is,$1.$

Step-by-step solution

Step 1. Given information

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

Step 2. Taking logarithm of the limit.

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

$=\underset{x\to 2^{+}}{\lim }\frac{\ln (x-2)}{\left(x^2-4\right)}$ [ in the form of $\frac{\infty }{\infty }$]

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

Step 3. Therefore, the value of the limit is given by,

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }\ln (x-2{)}^{x^2-4}=e^0 \\ =1 \end{gathered}$$

Therefore, the exact value is,$1.$