Chapter 11.3, Problem 80E

Solution

a.

To draw: The graph of the function $f(x)=\left\{\begin{array}{ll}\sin x\hfill & \text{ if }-\pi \le x<0\hfill \\ \csc x\hfill & \text{ if }0\le x\le \pi \hfill \end{array}\right.$ .

Given information:

The function $f(x)=\left\{\begin{array}{ll}\sin x\hfill & \text{ if }-\pi \le x<0\hfill \\ \csc x\hfill & \text{ if }0\le x\le \pi \hfill \end{array}\right.$ .

Graph:

Create a table of values for the given function as:

  

Now, plot these points and join them with a smooth curve.

  

b.

The points $c$ in the domain of the function at which does $\underset{x\to c}{\lim }f(x)$ exist.

The limit $\underset{x\to c}{\lim }f(x)$ exists at all the points except $x=-\pi ,0,\pi$ .

Given information:

The function $f(x)=\left\{\begin{array}{ll}\sin x\hfill & \text{ if }-\pi \le x<0\hfill \\ \csc x\hfill & \text{ if }0\le x\le \pi \hfill \end{array}\right.$ .

Calculation:

The function is not defined at the left of the point $x=-\pi$ and at the right of the point $x=\pi$ .

So, $\underset{x\to c}{\lim }f(x)$ does not exist at these two points.

Also, observe that the function approach 0 from the left of $x=0$ and approaches $\infty$ from the right of the point $x=0$ . Thus, $\underset{x\to 0}{\lim }f(x)$ does not exist.

Therefore, $\underset{x\to c}{\lim }f(x)$ exists at all the points except $x=-\pi ,0,\pi$ .

Conclusion:

The limit $\underset{x\to c}{\lim }f(x)$ exists at all the points except $x=-\pi ,0,\pi$ .

c.

The points $c$ in the domain of the function at which only left-hand limit exists.

The point at which the left-hand limit exists is $c=\pi$ .

Given information:

The function $f(x)=\left\{\begin{array}{ll}\sin x\hfill & \text{ if }-\pi \le x<0\hfill \\ \csc x\hfill & \text{ if }0\le x\le \pi \hfill \end{array}\right.$ .

Calculation:

Observe that the function is only defined at the left of the point $x=\pi$ , and the function approaches $\infty$ from the left at this point. Apart from that the right-hand limit exists at all the points in the domain of the function.

So, the only point at which the left-hand limit exists is $c=\pi$ .

Conclusion:

The point at which the left-hand limit exists is $c=\pi$ .

d.

The points $c$ in the domain of the function at which only right-hand limit exists.

The point at which the right-hand limit exists is $c=-\pi$ .

Given information:

The function $f(x)=\left\{\begin{array}{ll}\sin x\hfill & \text{ if }-\pi \le x<0\hfill \\ \csc x\hfill & \text{ if }0\le x\le \pi \hfill \end{array}\right.$ .

Calculation:

Observe that the function is only defined at the right of the point $x=-\pi$ , and the function approaches 0 from the right at this point. Apart from that the left-hand limit exists at all the points in the domain of the function.

So, the only point at which the right-hand limit exists is $c=-\pi$ .

Conclusion:

The point at which the right-hand limit exists is $c=-\pi$ .