Precalculus - 12.2a Techniques for Evaluating Limits

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 Substitution

lim (.5x³ - 5x) = (.5)(.2)² - 5(2) = 6
^x®-2

Example 2:
lim (x² + 1)/(x) = (3² + 1)/ 3 = (9 + 1)/ 3 = 10/3
^x®3

Example 3:
lim (x² - 1)^1/3 = (3² - 1)^1/3 = (9 - 1)^1/3 = 8^1/3 = 2
^x®3

II. Factoring: Dividing out Technique

Example 4:
lim (x² - 3x)/x = (x)(x-3)/(x)
^x®0

lim x - 3 = -3
^x®0

Example 5:
lim (x² - 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 » e²

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