Chapter 2, Problem 9RE

To determine

To find: The value of given limit.

Answer to Problem 9RE

The value of $\underset{x\to 0}{\lim }\frac{x\csc x+1}{x\csc x}$ is $2$ .

Explanation of Solution

Given information: The given expression is $\underset{x\to 0}{\lim }\frac{x\csc x+1}{x\csc x}$ .

Calculation:

To find the limit, simplify the function.

  $\begin{array}{c}\frac{x\csc x+1}{x\csc x}=\frac{x\csc x}{x\csc x}+\frac{1}{x\csc x}\\ =1+\frac{1}{\frac{x}{\sin x}}\\ =1+\frac{\sin x}{x}\end{array}$

It is know that $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$ .

Solve for the limit.

  $\begin{array}{c}\underset{x\to 0}{\lim }\left(1+\frac{\sin x}{x}\right)=\underset{x\to 0}{\lim }1+\underset{x\to 0}{\lim }\frac{\sin x}{x}\\ =1+1\\ =2\end{array}$

Therefore, the value of $\underset{x\to 0}{\lim }\frac{x\csc x+1}{x\csc x}$ is $2$ .