Chapter 9.2, Problem 21E

To determine

To Find: The value of a limit using L'Hôpital's Rule, after identifying the indeterminate form.

Answer to Problem 21E

The value of the limit is $e^2$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{\frac{1}{x}}$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $a\to 0$ ,

  $e^a\to 1,a\to a$ and $\frac{1}{a}\to \infty$

Therefore, the limit is of the indeterminate form $1^{\infty }$ .

So, assume the limit is $L$ and take the natural logarithm of the equation $L=\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{\frac{1}{x}}$ so that

  $\begin{array}{l}L=\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{\frac{1}{x}}\\ \Rightarrow \ln L=\ln \left[\underset{x\to 0}{\lim }{\left(e^x+x\right)}^{\frac{1}{x}}\right]\\ \Rightarrow \ln L=\underset{x\to 0}{\lim }\left[\ln {\left(e^x+x\right)}^{\frac{1}{x}}\right]\\ \Rightarrow \ln L=\underset{x\to 0}{\lim }\left[\frac{\ln \left(e^x+x\right)}{x}\right]\end{array}$

Now, the righthand side of the above equation is in $\frac{0}{0}$ form.

So, apply L'Hôpital's Rule on the righthand side of the equation to write

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

Therefore, the value of the limit is $e^2$ .