**12.2 Techniques for Evaluating Limits**

**I. Limits of Polynomial and Rational Functions**

1. If p is a polynomial function and "c" is a real number, then

lim p(x) = p(c)

^{x®c}

2. If "r" is a rational function given by r(x) = p(x)/q(x) and "c" is a real number such that q(x) ¹0.

**Example 1: Evaluating Limits by Direct Substituti**on

lim (.5x

^{3}- 5x) = (.5)(.2)

^{2}- 5(2) = 6

^{x®-2}

**Example 2**:

lim (x

^{2}+ 1)/(x) = (3

^{2}+ 1)/ 3 = (9 + 1)/ 3 = 10/3

^{x®3}

**Example 3:**

lim (x

^{2 }- 1)

^{1/3}= (3

^{2}- 1)

^{1/3}= (9 - 1)

^{1/3}= 8

^{1/3}= 2

^{x®3}

**II. Factoring: Dividing out Technique**

**Example 4:**

lim (x

^{2}- 3x)/x = (x)(x-3)/(x)

^{x®0}

lim x - 3 = -3

^{x®0}

**Example 5:**

lim (x

^{2}- 1)/(x + 1) =

^{x®-1}

lim(x + 1)(x - 1)/(x + 1) =

^{x®-1}

lim (x - 1) = -2

^{x®-1}

This technique should be applied only when direct substitution produces zero in both the numerator and the denominator. The resulting fraction 0/0, has no meaning as a real number.

It is called an indeterminate form because you cannot, from the form alone, determine the limit.

**III. Rationalizing Technique**

First Rationalize the numerator of the function.

**Example 6**:

lim (3 - Ö (x))/(x - 9)

^{x®9}

multiply the numerator and the denominator by the conjugate of the numerator:

lim[ (3 - Ö (x))(3 + Ö(x))] /[(x - 9) (3 + Ö(x))] =

^{x®9}

lim (9 - x) / [(x - 9 )(3 + Ö(x))] =

^{x®9}

As you can see in the numerator, you have (9 - x) and in the denominator you have (x - 9) so if we multiply the numerator by (-1) you would have:

lim (-1)(- 9 + x) / [(x - 9 )(3 + Ö(x))] =

^{x®9}

so now x - 9 divides out of the numerator and denominator

lim -1 / (3 + Ö(9)) = -1/ (3 + 3 ) = -1/6

^{x®9}

**IV. Using Technology**:

lim (1 + 2x)

^{1/x}

^{x®0}

Since you cannot use substitution, use a table:

(x, f(x)) = {(-.1, 9.313), (-.01, 7.540), (-.001, 7.404), (-.0001, 7.391), (-.00001, 7.3892), (-.000001, 7.38907), (.000001, 7.38904), (.00001, 7.3898), etc...}

so you can see that it is estimated about

lim (1 + 2x)

^{1/x}»7.3890561 » e

^{2}

^{x®0}

**Algebraically:**

Choose 2 points that are equidistant from 0 and close to 0 and find their average:

(7.38907 + 7.38904)/2 »7.389055 » e

^{2}

**Example 7:**

lim (1/(2+x) - 1/2)/x =

^{x®0}

again using the calculator when x = -.0000000001, f(x) = -1/4

and when x = .0000000001, f(x) = -1/4

so

lim (1/(2+x) - 1/2)/x = -1/4

^{x®0}

**Algebraically**:

lim (1/(2+x) - 1/2)/x =

^{x®0}

simplifying the fraction:

[(2 - (2 + x))/(2 + x)(2)]/x =

(- x)/ [(2 + x)(2)(x)] =

-1/(4 + 2x) so:

lim (1/(2+x) - 1/2)/x = -1/(4 + 2(0)) = -1/4 same answer!!

^{x®0}