Chapter 1.3, Problem 60E

(a)

To determine

To find : the domain of the function $f$ .

Answer to Problem 60E

The domain of the function is $\left(-\infty ,-1\right)\cup \left(0,\infty \right)$ .

Explanation of Solution

Given information :

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

Calculation :

Substitute $x=0$ in the function $f\left(x\right)={\left(1+\frac{1}{x}\right)}^x$ .

  $\begin{array}{c}f\left(0\right)={\left(1+\frac{1}{0}\right)}^0\\ ={\left(\frac{1}{0}\right)}^0\end{array}$

The function is not defined at $x=0$ .

Substitute $x=-1$ in the function $f\left(x\right)={\left(1+\frac{1}{x}\right)}^x$ .

  $\begin{array}{c}f\left(0\right)={\left(1+\frac{1}{-1}\right)}^{-1}\\ ={\left(0\right)}^{-1}\\ =\infty \end{array}$

The function is not defined at $x=-1$ .

So, the function is not defined in the interval $\left[-1,0\right]$ .

Therefore, the domain of the function is $\left(-\infty ,-1\right)\cup \left(0,\infty \right)$ .

(b)

To determine

To graph : The function $f$ .

Explanation of Solution

Given information :

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

Graph :

Thus, the graph is

  

Interpretation :

One vertical asymptote at $x=-1$ and not defined at $x=0$ .

(c)

To determine

To explain : the point of discontinuity of the function $f$ .

Explanation of Solution

Given information :

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

Calculation :

  

On observing the graph, vertical asymptote at $x=-1$ and not defined at $x=0$ .

That why $x=-1$ and $x=0$ are the points of discontinuity.

(d)

To determine

To explain : the point of discontinuity is removable or not.

Explanation of Solution

Given information :

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

Calculation :

On observing the graph, vertical asymptote at $x=-1$ .

The discontinuity at $x=-1$ is not removable because it has infinite discontinuity at this point.

Function not defined at $x=0$ .

The discontinuity at  $x=0$  is also not removable. If set  $\underset{x\to 0^{+}}{\lim }f\left(x\right)=1$ , the $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ does not exist. Thus, limit itself does not exist.

(e)

To determine

To estimate : the value of the function as $x$ tends to infinity.

Answer to Problem 60E

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

Explanation of Solution

Given information :

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

Calculation :

Make data table for given function.

  $\begin{array}{cccccc}x& 10& 100& 1000& 10000& 100000\\ f\left(x\right)& 2.5937& 2.70481& 2.71692& 2.71814& 2.71827\end{array}$

As the value of $x$ increases, the value of function gets constant and approximate to $e$ .

Hence, $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e$ .