Chapter 9.2, Problem 9E

(a).

To determine

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

Answer to Problem 9E

The value of the limit is $2$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{-}}{\lim }\frac{\sin 4x}{\sin 2x}$

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, from both the sides, the value of the expression approaches to $2$ near $x=0$ .

Therefore, the value of the limit is $2$ .

(ii). Apply L'Hôpital's Rule to write

  $\underset{x\to 0^{-}}{\lim }\left(\frac{\sin 4x}{\sin 2x}\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{4\cos 4x}{2\cos 2x}\right)=\frac{4\cos 0^{-}}{2\cos 0^{-}}=\frac{4\cdot 1}{2\cdot 1}=2$

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 9E

The value of the limit is $2$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{+}}{\lim }\frac{\sin 4x}{\sin 2x}$

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, from both the sides, the value of the expression approaches to $2$ near $x=0$ .

Therefore, the value of the limit is $2$ .

(ii). Apply L'Hôpital's Rule to write

  $\underset{x\to 0^{+}}{\lim }\left(\frac{\sin 4x}{\sin 2x}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{4\cos 4x}{2\cos 2x}\right)=\frac{4\cos 0^{+}}{2\cos 0^{+}}=\frac{4\cdot 1}{2\cdot 1}=2$

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