Chapter 11.2, Problem 85E

a.

To determine

To give conclusion of the given limit.

Explanation of Solution

Given:

  $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{0}{1}$

Conclusion:

If $\frac{p(x)}{q(x)}$ is a rational function and $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=0$ then ,

If p (c) =0 it could be cancel the factor of a polynomial of p (x)

With a factor of a polynomial q (x) and a resulting rational function

has no limit at x = c

b.

To determine

To give conclusion of the given limit.

Explanation of Solution

Given:

  $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{1}{1}$

Conclusion:

If $\frac{p(x)}{q(x)}$ is a rational function and $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{1}{1}$ then ,

If p (c) =1 it could be cancel the factor of a polynomial of p (x)

With a factor of a polynomial q (x) and a resulting rational function

has finite limit.

c.

To determine

To give conclusion of the given limit

Explanation of Solution

Given:

  $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{1}{0}$

Conclusion:

If $\frac{p(x)}{q(x)}$ is a rational function and $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{1}{0}$ then ,

Limit does not exist because 0 in the denominator results undefined.

d.

To determine

To give conclusion of the given limit

Explanation of Solution

Given:

  $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{0}{0}$

..

Conclusion:

If $\frac{p(x)}{q(x)}$ is a rational function and $\underset{x\to c}{\lim }\frac{p(x)}{q(x)}=\frac{0}{0}$ then ,

If f (x) is a rational function where p (x) and q (x) are polynomials with q (c) =0 and

p (c) = 0 we can cancel the factor of a polynomial of p (x) With a factor of a polynomial

q (x) and a resulting rational function has no limit at x = c