Chapter 1.1, Problem 77E

(a.)

To determine

To explain: Why limit is restricted to $0<\theta <\frac{\pi }{2}$ in order to show that the right-hand limit is 1. The given figure is shown below:

  

Answer to Problem 77E

From the figure it can be observed that the graph of the function lies between $0<\theta <\frac{\pi }{2}$ and right-hand limit when x approaching to 0 is depends only for the positive x values near zero.

Explanation of Solution

Given:

The given figure is shown below:

  

Consider the given graph.

It is known that an output value that a function approaches for the specified input values is referred to as a limit.

From the given figure it can be observed that the value of the function or the limit of the function when x approaching to 0 is depends only for the positive x values near zero. This is why limit is restricted to $0<\theta <\frac{\pi }{2}$ .

(b.)

To determine

To Prove: The area of $\Delta OAP=\frac{1}{2}\sin \theta$ , the area of sector $OAP=\frac{\theta }{2}$ and the area of

  $\Delta OAT=\frac{1}{2}\tan \theta$ .

Explanation of Solution

Given:

The given figure is shown below:

  

Concept used:

Area of triangle is: $A=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)$

Area of sector is: $=\frac{1}{2}\left(\text{angle}\right){\left(\text{radius}\right)}^2$

Calculation:

Consider the given figure.

From the figure it can be observed that, in $\Delta OAP$ , base is $OA=1$ and height is $PQ=\sin \theta$ .

Put $\text{Base}=1$ and $\text{height}=\sin \theta$ into the area of triangle formula.

  $\begin{array}{c}\Delta OAP=\frac{1}{2}\left(\text{1}\right)\left(\text{sin}\theta \right)\\ =\frac{1}{2}\sin \theta \end{array}$

Consider the sector $OAP$ .

From the figure it can be observed that, in sector $OAP$ angle is $\theta$ and radius is 1 unit.

Use the area of the sector formula:

  $\begin{array}{c}A=\frac{1}{2}\left(\text{angle}\right){\left(\text{radius}\right)}^2\\ =\frac{1}{2}\left(\theta \right){\left(\text{1}\right)}^2\\ =\frac{\theta }{2}\end{array}$

Consider the triangle $\Delta OAT$ .

From the figure it can be observed that, in $\Delta OAT$ , base is $OA=1$ and height is $AT=\tan \theta$ .

Put $\text{Base}=1$ and $\text{height}=\tan \theta$ into the area of triangle formula.

  $\begin{array}{c}\Delta OAT=\frac{1}{2}\left(\text{1}\right)\left(\tan \theta \right)\\ =\frac{1}{2}\tan \theta \end{array}$

(c.)

To determine

To Prove: That $\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta$ , for $0<\theta <\frac{\pi }{2}$ by using the part (b) and the figure.

Explanation of Solution

Given:

From part (b) The area of $\Delta OAP=\frac{1}{2}\sin \theta$ , the area of sector $OAP=\frac{\theta }{2}$ and the area of

  $\Delta OAT=\frac{1}{2}\tan \theta$ .

The given figure is shown below:

  

Concept used:

In a right-angled triangle one of its angles is $90°$ .

Calculation:

Consider the given information.

The triangle given in the figure is right angle triangle and in a right angle triangle one angle is $90°$ .

Also, it is known that the area of $\Delta OAP=\frac{1}{2}\sin \theta$ , the area of sector $OAP=\frac{\theta }{2}$ and the area of

  $\Delta OAT=\frac{1}{2}\tan \theta$ .

Now observe the value of $\sin \theta$ is lies between $-1$ to 1. The value of $\theta$ is $0<\theta <\frac{\pi }{2}$ and the value of $\tan \theta$ is lies between $-\infty$ to $\infty$ .

Also, from the figure it can be observed that $\text{Area of }\Delta OAP<\text{ area of sector }OAP$ .

Therefore, $\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta$ .

(d.)

To determine

To Prove: That $\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta$ can be written as $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$ for $0<\theta <\frac{\pi }{2}$ .

Explanation of Solution

Given:

The given inequality is: $\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta$

Concept used:

Domain of a function is the set of input values which a function can take.

Calculation:

Consider the given inequality.

  $\frac{1}{2}\sin \theta <\frac{1}{2}\theta <\frac{1}{2}\tan \theta$

Mulitiply each side of the inequality by 2.

  $\begin{array}{c}2\times \frac{1}{2}\sin \theta <2\times \frac{1}{2}\theta <2\times \frac{1}{2}\tan \theta \\ \sin \theta <\theta <\tan \theta \end{array}$

Now divide each side by $\sin \theta$ . Since $\sin \theta$ is an increasing function in domain $0<\theta <\frac{\pi }{2}$ so don’t change the inequality sign.

  $\begin{array}{c}\frac{\sin \theta }{\sin \theta }<\frac{\theta }{\sin \theta }<\frac{\tan \theta }{\sin \theta }\\ 1<\frac{\theta }{\sin \theta }<\frac{\sin \theta }{cos\theta \times \sin \theta }\\ 1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }\end{array}$

(e.)

To determine

To Prove: That $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$ can be written as $\cos \theta <\frac{\sin \theta }{\theta }<1$ for $0<\theta <\frac{\pi }{2}$ .

Explanation of Solution

Given:

The given inequality is: $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$

Concept used:

Domain of a function is the set of input values which a function can take.

Calculation:

Consider the given inequality.

  $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$

Taking reciprocal of the above inequality and change the direction of the inequlaity.

  $\begin{array}{c}1>\frac{\sin \theta }{\theta }>\cos \theta \\ \cos \theta <\frac{\sin \theta }{\theta }<1\end{array}$

(f.)

To determine

To Prove: $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }=1$ by using the squeeze theorem.

Explanation of Solution

Given:

The given limit is $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }$

Concept used:

The squeeze theorem states that if $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all numbers, and at some point $x=k$ we have $f\left(k\right)=h\left(k\right)$ , then $g\left(k\right)$ must also be equal to them. 

Calculation:

Consider the given information.

From part (e) it is known that $\cos \theta <\frac{\sin \theta }{\theta }<1$ for $0<\theta <\frac{\pi }{2}$

Applying limit on each function.

Let $f\left(x\right)=\underset{\theta \to 0^{+}}{\lim }\cos \theta$

Thus, $\underset{\theta \to 0^{+}}{\lim }\cos \theta =1$

Let $h\left(x\right)=\underset{\theta \to 0^{+}}{\lim }1$

Thus, $\underset{\theta \to 0^{+}}{\lim }1=1$

Therefore, by the squeeze theorem $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }=1$ because in the given interval the limits of both $\cos \theta$ and 1 is 1.

(g.)

To determine

To Prove: That $\frac{\sin \theta }{\theta }$ is an even function.

Explanation of Solution

Given:

The given function is $\frac{\sin \theta }{\theta }$ .

Concept used:

A function is said to be even if $f\left(-x\right)=f\left(x\right)$

Calculation:

Consider the given function.

  $f\left(\theta \right)=\frac{\sin \theta }{\theta }$

Replace $\theta$ with $-\theta$ .

  $\begin{array}{c}f\left(-\theta \right)=\frac{\sin \left(-\theta \right)}{\left(-\theta \right)}\\ =\frac{-\sin \theta }{-\theta }\\ =\frac{\sin \theta }{\theta }\\ =f\left(\theta \right)\end{array}$

Since $f\left(-\theta \right)=f\left(\theta \right)$ thus the given function is an even function.

(h.)

To determine

To Prove: That $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$ by using the result obtained in part (g).

Explanation of Solution

Given:

From part (g) it is known that $\frac{\sin \theta }{\theta }$ is an even function.

Concept used:

The squeeze theorem states that if $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ for all numbers, and at some point $x=k$ we have $f\left(k\right)=h\left(k\right)$ , then $g\left(k\right)$ must also be equal to them. 

Calculation:

Consider the given limit.

  $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$

Take $\theta =0-h,h\to 0$

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{\sin \left(0-h\right)}{\left(0-h\right)}=\underset{h\to 0}{\lim }\frac{\sin \left(-h\right)}{\left(-h\right)}\\ =\underset{h\to 0}{\lim }\frac{-\sin \left(h\right)}{\left(-h\right)}\\ =\underset{h\to 0}{\lim }\frac{\sinh }{h}\\ =1\end{array}$

Therefore, $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$ .

(i.)

To determine

To Prove: $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ .

Explanation of Solution

Given:

From part (f) and (h) it is known that $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }=1$ and $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$ respectively.

Concept used:

The limit of a function as it approaches from the left is referred to as the left-hand limit. The limit of a function as it approaches from the right-hand side is referred to as a right-hand limit.

Calculation:

Consider the given information.

From part (f) and (h) it is known that $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }=1$ and $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$ respectively.

  $\underset{\theta \to 0^{+}}{\lim }\frac{\sin \theta }{\theta }=1$ represents the right-hand limit and $\underset{\theta \to 0^{-}}{\lim }\frac{\sin \theta }{\theta }=1$ represents the left-hand limit.

If the function has both limits defined at a particular x value c and those values match, then the limit will exist and will be equal to the value of the one-sided limits.

Since the left- and right-hand limits are equal so $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ .

Therefore, $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ .