Chapter 9.2, Problem 25E

To determine

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

Answer to Problem 25E

The value of the limit is $1$ .

Explanation of Solution

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

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

Calculation: As $a\to 0^{+}$ ,

  $1+\frac{1}{a}\to \infty$ and $a\to 0$

Therefore, the limit is of the indeterminate form ${\infty }^0$ .

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

  $\begin{array}{l}L=\underset{x\to 0^{+}}{\lim }{\left(1+\frac{1}{x}\right)}^x\\ \Rightarrow \ln L=\ln \left[\underset{x\to 0^{+}}{\lim }{\left(1+\frac{1}{x}\right)}^x\right]\\ \Rightarrow \ln L=\underset{x\to 0^{+}}{\lim }\left[\ln {\left(1+\frac{1}{x}\right)}^x\right]\\ \Rightarrow \ln L=\underset{x\to 0^{+}}{\lim }\left[x\ln \left(1+\frac{1}{x}\right)\right]\\ \Rightarrow \ln L=\underset{x\to 0^{+}}{\lim }\frac{\ln \left(1+\frac{1}{x}\right)}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\end{array}$

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

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

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

Therefore, the value of the limit is $1$ .