Chapter 2.2, Problem 10E

(a)

To determine

To estimate: The value of $\underset{x\to 0^{-}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}$ from the graph of $f\left(x\right)=\frac{x^2+x}{\sqrt{x^3+x^2}}$.

Answer to Problem 10E

The value of $\underset{x\to 0^{-}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=-1$.

Explanation of Solution

Using the graphing calculator, the graph of the function $f\left(x\right)=\frac{x^2+x}{\sqrt{x^3+x^2}}$ is drawn and shown below in Figure 1.

From Figure 1, it is observed that the graph approaches $y=-1$ as x approaches 0 from the left side.

Thus, the value of $\underset{x\to 0^{-}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=-1$.

(b)

To determine

To estimate: The value of $\underset{x\to 0^{+}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}$ from the graph of $f\left(x\right)=\frac{x^2+x}{\sqrt{x^3+x^2}}$.

Answer to Problem 10E

The value of $\underset{x\to 0^{+}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=1$.

Explanation of Solution

From Figure 1, it is observed that the graph approaches $y=1$ as x approaches 0 from the right side.

Thus, the value of $\underset{x\to 0^{+}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=1$.

(c)

To determine

To estimate: The value of $\underset{x\to 0}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}$ from the graph of $f\left(x\right)=\frac{x^2+x}{\sqrt{x^3+x^2}}$.

Answer to Problem 10E

The value of $\underset{x\to 0}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}$ does not exist.

Explanation of Solution

From Figure 1, it is observed that the graph does not approach any value when x equals exactly 0.

Also, notice the left hand limit and the right limit of $\underset{x\to 0}{\lim }\frac{1}{1+e^{\frac{1}{x}}}$ obtained from part (a) and part (b) are not the same.

That is $\underset{x\to 0^{-}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=-1$ and $\underset{x\to 0^{+}}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}=1$.

Thus, it is concluded that $\underset{x\to 0}{\lim }\frac{x^2+x}{\sqrt{x^3+x^2}}$ does not exist.