Chapter 12.2, Problem 40E

(a)

To determine

Whichlimit can be evaluated by using direct substitution. Thenevaluate or approximate each limit.

Answer to Problem 40E

This limit can be evaluated using direct substitution. Value of $\underset{x\to 0}{\lim }{x}^2\sin x^2=0$

Explanation of Solution

Given info:

  $\underset{x\to 0}{\lim }\frac{x}{\cos x}$

Formula used:

If f(x) becomes arbitrarily close to a unique number L as approaches c fromeither side, then the limit of f(x) as approaches c is L. This is written as

  $\underset{x\to c}{\lim }f(x)=L$

Calculation:

We have

  $\underset{x\to 0}{\lim }\frac{x}{\cos x}$

Substituting directly,

  $\underset{x\to 0}{\lim }\frac{x}{\cos x}=\frac{0}{\cos 0}=0$

This limit can be evaluated using direct substitution. Value of $\underset{x\to 0}{\lim }\frac{x}{\cos x}=0$

Conclusion:

Thus,this limit can be evaluated using direct substitution. Value of $\underset{x\to 0}{\lim }\frac{x}{\cos x}=0$

(b)

To determine

Whichlimit can be evaluated by using direct substitution. Thenevaluate or approximate each limit.

Answer to Problem 40E

This limit cannot be evaluated using direct substitution. Value of $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}$

Explanation of Solution

Given info:

  $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}$

Formula used:

If f(x) becomes arbitrarily close to a unique number L as approaches c fromeither side, then the limit of f(x) as approaches c is L. This is written as

  $\underset{x\to c}{\lim }f(x)=L$

Calculation:

We have

  $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}$

Substituting directly,

  $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1-\cos 0}{0^2}=\frac{0}{0}$

This limit cannot be evaluated using direct substitution.

Value of

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\underset{x\to 0}{\lim }\frac{2{\sin }^2\left(\frac{x}{2}\right)}{x^2}=2\underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{2}\right)}{x}\times \frac{\sin \left(\frac{x}{2}\right)}{x}\\ \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=2\underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{2}\right)}{2\times \frac{x}{2}}\times \underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{2}\right)}{2\times \frac{x}{2}}\\ \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{2}{4}\underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\times \underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\\ \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}\times 1\times 1\\ \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}\end{array}$

Conclusion:

Thus,this limit cannot be evaluated using direct substitution. Value of $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}$