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

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

Short answer

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

Step-by-step solution

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:

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

Therefore,the value of liimit$1$