Chapter 9.2, Problem 12E

(a).

To determine

To Find: The value of a one-sided limit, both graphically and using L'Hôpital's Rule.

Answer to Problem 12E

The value of the limit is $-\infty$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{-}}{\lim }\frac{\tan x}{x^2}$

Method:

Graphically: : Plot the graph of the expression in $x$ and identify the value to which the expression approaches as $x$ approaches to $0$ from the left.

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i)The graph of the expression is

  

In the above graph, the value of the expression approaches to $-\infty$ when $x$ approaches to $0$ from the left.

Therefore, the value of thelimit is $-\infty$ .

(ii). Apply L'Hôpital's Rule to write

  $\underset{x\to 0^{-}}{\lim }\left(\frac{\tan x}{x^2}\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{{\sec }^{2}x}{2x}\right)=\frac{{\sec }^2{0}^{-}}{2\times 0^{-}}=\frac{1}{2\times (0^{-})}=-\infty$

The limit determined using both the methods is indeed the same.

(b).

To determine

To Find: The value of a one-sided limit, both graphically and using L'Hôpital's Rule.

Answer to Problem 12E

The value of the limit is $\infty$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x^3}$

Method:

Graphically: : Plot the graph of the expression in $x$ and identify the value to which the expression approaches as $x$ approaches to $0$ from the right.

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i)The graph of the expression is

  

In the above graph, the value of the expression approaches to $\infty$ when $x$ approaches to $0$ from the right.

Therefore, the value of thelimit is $\infty$ .

(ii). Apply L'Hôpital's Rule to write

  $\underset{x\to 0^{+}}{\lim }\left(\frac{\tan x}{x^2}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{{\sec }^{2}x}{2x}\right)=\frac{{\sec }^2{0}^{+}}{2\times 0^{+}}=\frac{1}{2\times 0^{+}}=\infty$

The limit determined using both the methods is indeed the same.