Chapter 15.1, Problem 1GCE

To determine

To find:

The value of given limit at that point .

Answer to Problem 1GCE

  $\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$

Explanation of Solution

Given:

The function $f$ is

  $\underset{x\to 0}{\lim }\frac{e^x-1}{x}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

  $\underset{x\to a}{\lim }f\left(x\right)=L$

Calculation:

The existence of the limit

  $\underset{x\to 0}{\lim }\frac{e^x-1}{x}$

Apply by limit

  $\begin{array}{l}=\frac{e^0-1}{0}\\ =\frac{0}{0}\end{array}$

Applying by L’ Hospital rule

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{e^x-1}{x}\\ \underset{x\to 0}{\lim }\frac{e^x-0}{1}\\ \underset{x\to 0}{\lim }{e}^x\end{array}$

Applying by limit

  $\begin{array}{l}=\underset{x\to 0}{\lim }{e}^x\\ =e^0\\ =1\end{array}$

The value of $\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$ .