Chapter 12.2, Problem 42E

a.

To determine

To evaluate:Thelimit of the given function by substitution method.

Answer to Problem 42E

Limit of the function is $0$ .

Explanation of Solution

Given:

  ${\underrightarrow{\lim }}_{x\to 0}\frac{x}{cosx}$

Explanation:Let

  $f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{x}{cosx}$ be a given function.

It has to find limit by substitution method.

  ${\underrightarrow{\lim }}_{x\to a}f\left(x\right)=f\left(a\right)$

  $\begin{array}{l}f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{x}{cosx}\\ f\left(0\right)=\frac{0}{cos0}\\ =0\end{array}$

Hence, limit of the function ${\underrightarrow{\lim }}_{x\to 0}\frac{x}{cosx}=0$ .

b.

To determine

To evaluate: The limit of the given function by substitution method.

Answer to Problem 42E

Limit of the function is $\frac{1}{2}$ .

Explanation of Solution

Given:

  ${\underrightarrow{\lim }}_{x\to 0}\frac{1-cosx}{x^2}$

Explanation: Let

  $f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{1-cosx}{x^2}$ be a given function.

It has to find limit by substitution method.

  ${\underrightarrow{\lim }}_{x\to a}f\left(x\right)=f\left(a\right)$

  $\begin{array}{l}f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{1-cosx}{x^2}\\ f\left(0\right)=\frac{0}{0}\left[\text{indeterminant form}\right]\end{array}$

Using L.Hospital rule $\begin{array}{l}f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{\sin x}{2x}\\ f\left(0\right)=\frac{0}{0}\left[\text{indeterminant form}\right]\end{array}$

Again using L.Hospital rule $\begin{array}{l}f\left(x\right)={\underrightarrow{\lim }}_{x\to 0}\frac{cosx}{2}\\ f\left(0\right)=\frac{cos0}{2}\\ =\frac{1}{2}\end{array}$

Hence, limit of the function ${\underrightarrow{\lim }}_{x\to 0}\frac{1-cosx}{x^2}=\frac{1}{2}$ .