Chapter 8.2, Problem 21E

To determine

To calculate: The limit of the given fraction.

Answer to Problem 21E

The obtained value is $e^2$ .

Explanation of Solution

Given Information:

The rational expression is defined as,

  $\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{1/x}$

Formula Used:

L’hospital rule is defined as,

If the limit is giving undefined form $\frac{\infty }{\infty },\frac{0}{0}$ for the function $\frac{f\left(x\right)}{g\left(x\right)}$ .

  $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\frac{d}{dx}\left(f\left(x\right)\right)}{\frac{d}{dx}\left(g\left(x\right)\right)}$

Calculation:

Consider the given information,

  $\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{1/x}$

Simplify the expression to apply the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{1/x}=\underset{x\to 0}{\lim }{e}^{\ln {\left(e^x+x\right)}^{1/x}}\\ =\underset{x\to 0}{\lim }{e}^{\frac{1}{x}\ln \left(e^x+x\right)}\\ =e^{\underset{x\to 0}{\lim }\frac{1}{x}\ln \left(e^x+x\right)}\end{array}$

Now, calculate the value of $\underset{x\to 0}{\lim }\frac{1}{x}\ln \left(e^x+x\right)$ with the help of L’hospital rule.

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{1}{x}\ln \left(e^x+x\right)=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\ln \left(e^x+x\right)}{\frac{d}{dx}x}\\ =\underset{x\to 0}{\lim }\frac{\frac{e^x+1}{e^x+x}}{1}\\ =\frac{e^0+1}{e^0+0}\\ =2\end{array}$

Rewrite the value.

  $\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{1/x}=e^2$

Hence, the required value is $e^2$ .