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

$\underset{x\to 2}{\lim }\frac{e^{2x-4}-1}{x^2-4}$

Short answer

The value of$\underset{x\to 2}{\lim }\frac{e^{2x-4}-1}{x^2-4}=\frac{1}{2}$

Step-by-step solution

Step 1. Given information

The given limit$\underset{x\to 2}{\lim }\frac{e^{2x-4}-1}{x^2-4}$

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

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{e^{2x-4}-1}{x^2-4}=\underset{x\to 2}{\lim }\frac{e^{2x-4}(2-0)}{2x-0} \\ =\underset{x\to 2}{\lim }\frac{2e^{2x-4}}{2x} \\ =\frac{2e^{2.2-4}}{2.2} \\ =\frac{e^0}{2} \\ =\frac{1}{2} \end{gathered}$$

Therefore,the value of given limit$=\frac{1}{2}$