Chapter 11.3, Problem 33E

Solution

a)

To find: $\underset{x\to 0^{-}}{\lim }f\left(x\right)$ where $f\left(x\right)={\left(1+x\right)}^{1/x}$ from the graph.

  $\underset{x\to -0^{-}}{\lim }f\left(x\right)\approx 2.7$

Given information:

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

Calculation:

Draw the graph of $f\left(x\right)={\left(1+x\right)}^{1/x}$ .

  

From the graph, it is evident that $\underset{x\to -0^{-}}{\lim }f\left(x\right)\approx 2.7$

b)

To find: $\underset{x\to 0^{+}}{\lim }f\left(x\right)$ where $f\left(x\right)={\left(1+x\right)}^{1/x}$ from the graph.

  $\underset{x\to -0^{+}}{\lim }f\left(x\right)\approx 2.7$

Given information: $f\left(x\right)={\left(1+x\right)}^{1/x}$

Calculation:

Draw the graph of $f\left(x\right)={\left(1+x\right)}^{1/x}$ .

  

From the graph, it is evident that $\underset{x\to -0^{-}}{\lim }f\left(x\right)\approx 2.7$

c)

To find: $\underset{x\to 0}{\lim }f\left(x\right)$ if it exists.

  $\underset{x\to 0}{\lim }f\left(x\right)\approx 2.7$ as $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)$ the function $f\left(x\right)={\left(1+x\right)}^{1/x}$ exists.

Given information:

Theorem states that a function $f\left(x\right)$ has a limit as $x$ approaches $c$ if and only if the left-hand and right-hand limits at $c$ exist and they are equal.

  $\underset{x\to c}{\lim }f\left(x\right)=L\iff \underset{x\to c^{-}}{\lim }f\left(x\right)=L$ and $\underset{x\to c^{+}}{\lim }f\left(x\right)=L$

Calculation:

Draw the graph of $f\left(x\right)={\left(1+x\right)}^{1/x}$ .

  

According to the theorem, a function is defined only if it’s left-hand and right-hand limits are equal.

As $\underset{x\to 0^{-}}{\lim }f\left(x\right)\approx 2.7$ and $\underset{x\to 0^{+}}{\lim }f\left(x\right)\approx 2.7$

Therefore, $\underset{x\to 0}{\lim }f\left(x\right)\approx 2.7$ as $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)$ the function $f\left(x\right)={\left(1+x\right)}^{1/x}$ exists.