Q. 22 - 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

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

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

The limit using L’Hopitals rule is given below:

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x-1}{e^{x-1}-1}=\underset{x\to 1}{\lim }\frac{1-0}{e^{x-1}(1-0)-0} [u\sin g L\text{'}Hopitals rule] \\ =\underset{x\to 1}{\lim }\frac{1}{e^{x-1}} \\ =\frac{1}{e^{1-1}} \\ =\frac{1}{e^0} \\ =\frac{1}{1} \\ =1 \end{gathered}$$

Therefore,the value of liimit $1$