Chapter 12.2, Problem 22E

To determine

To find: The limit of given function and use graphing utility to verify your answer.

Answer to Problem 22E

Limit of the function ${\underrightarrow{\lim }}_{x\to \pi /2}\frac{1-\sin x}{\cos x}$ is $0$ .

Explanation of Solution

Given: ${\underrightarrow{\lim }}_{x\to \pi /2}\frac{1-\sin x}{\cos x}$

Consider $f\left(x\right)={\underrightarrow{\lim }}_{x\to \pi /2}\frac{1-\sin x}{\cos x}$ be a given function.

Now limit of the function is

  $f\left(x\right)={\underrightarrow{\lim }}_{x\to \pi /2}\frac{1-\sin x}{\cos x}$

After putting the value it gives indeterminate form, so it apply L. hospital rule

  $\begin{array}{l}={\underrightarrow{\lim }}_{x\to \pi /2}\frac{-\cos x}{-\sin x}\\ =\frac{\cos \left(\pi /2\right)}{\sin \left(\pi /2\right)}\end{array}$

  $=0$

And graph of this function is

  

So, the limit of the function ${\underrightarrow{\lim }}_{x\to \pi /2}\frac{1-\sin x}{\cos x}$ is $0$ .