Chapter 9.2, Problem 61E

a.

To determine

To find: The domain of function $f\left(x\right)$

Answer to Problem 61E

  $\therefore \text{ domain is }\left(-\infty ,-1\right)\cup \left(0,\infty \right)$

Explanation of Solution

Given information :

  $f\left(x\right)={\left(1+\frac{1}{x}\right)}^x$

Formula used : $a^x=e^{x\ln a}$

Calculation :

We have ${\left(1+\frac{1}{x}\right)}^x=e^{\ln {\left(1+\frac{1}{x}\right)}^x}=e^{\ln \left(1+\frac{1}{x}\right)}$

In this case $\left(1+\frac{1}{x}\right)$ should be greater than $0$ .

I.e.

  $\begin{array}{l}1+\frac{1}{x}\rangle 0\Rightarrow x^2\left(1+\frac{1}{x}\right)\rangle 0\times x^2\text{ as }{x}^2\ge 0\\ \Rightarrow x\left(1+x\right)\rangle 0\Rightarrow \text{either }x\rangle 0\text{ or }x\langle -1\end{array}$

  $\therefore \text{ domain is }\left(-\infty ,-1\right)\cup \left(0,\infty \right)$

b.

To determine

To find: $\underset{x\to -1^{-}}{\lim }f\left(x\right)$

Answer to Problem 61E

  $\underset{x\to -1^{-}}{\lim }f\left(x\right)=\infty$

Explanation of Solution

Given information :

  $\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{-}}{\lim }{\left(1+\frac{1}{x}\right)}^x$

Formula used : $a^x=e^{x\ln a}$

Calculation:

  $\begin{array}{c}\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{-}}{\lim }{\left(1+\frac{1}{x}\right)}^x\\ =\underset{x\to -1^{-}}{\lim }{\left(1+x^{-1}\right)}^x\\ =\underset{x\to -1^{-}}{\lim }{\left(1-1\right)}^{-1}\\ =\infty \end{array}$

c.

To determine

To Find: $\underset{x\to \infty }{\lim }f\left(x\right)$

Answer to Problem 61E

  $\underset{x\to \infty }{\lim }f\left(x\right)=1$

Explanation of Solution

Given information :

  $\underset{x\to \infty }{\lim }f\left(x\right)={\left(1+\frac{1}{x}\right)}^x$

Formula used : $a^x=e^{x\ln a}$

Calculation:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)={\left(1+\frac{1}{x}\right)}^x\\ =\underset{y\to \infty }{\lim }{\left(1+\frac{1}{y}\right)}^{-y}\left[y=-x\left(say\right)\right]\\ =\underset{y\to \infty }{\lim }\frac{1}{1-\frac{1}{y}}\\ =\frac{1}{1-0}=1\end{array}$