## Thursday, April 26, 2007

### 12.2b Precalculus - One-Sided Limits

12.2b One-Sided Limits

One-Sided Limits - when a limit approaches one value from the left side of "c" and a different value from the right side of "c"

lim f(x) = L
x®c-

and

lim f(x) = K
x®c+

where if L = K, then the limit exists but if L ¹K, then the limit does not exist for the value of x.

Example 1:

lim f(x)
x®1

Given f(x) = 4 - x2 when x£1 or f(x) = 3 - x when x is greater than 1

therefore:

lim f(x) = 4 - (1)2 = 4 - 1 = 3
x®1-

and

lim f(x) = 3 - 1 = 2
x®1+

so since 3 ¹ 2, then
lim f(x) does not exist since the function approaches different values
x®1
from the left and from the right.

I. Existence of a Limit:

If "f" is a function and "c" and "L" are real numbers, then

lim f(x) = L
x®c

if and only if BOTH the left and the right limits exist and are equal to L.

Example 2:

f(x) = 3 - x when x is less than 1 or 3x - x2 when x is greater than 1

lim f(x) = 2
x ®1-

lim f(x) = 2
x ®1+

Because the one-sided limits both exist and are equal to the same number 2, it follows that

lim f(x) = 2 so the limit does exist at x = 1.
x®1

II. A limit from Calculus:

Evaluating a Limit from Calculus - the limit of a difference quotient:

lim (f(x+h) - f(x))/h where h¹0
h®0

Direct substitution into the difference quotient ALWAYS produces the indeterminate
form 0/0.

Example 3: find the difference quotient of f(x) = 5 - 6x

lim [5 - 6(x + h) - (5 - 6x)]/h
h®0

lim [5 - 6x - 6h - 5 + 6x]/h
h®0

lim -6h/h = -6
h®0

therefore:
lim [5 - 6(x + h) - (5 - 6x)]/h = -6
h®0

Example 4: find the difference quotient of f(x)= 4 - 2x - x2

lim ([4 - 2(x + h) - (x+h)2] - [4 - 2x - x2])/h =
h®0

lim ([4 - 2x -2h -(x2 + 2xh + h2] - 4 + 2x + x2)/h =
h®0

lim (4 - 2x -2h -x2 -2xh - h2 - 4 + 2x + x2)/h =
h®0

lim (- 2h - 2xh - h2)/h =
h®0

lim (- 2 -2x -h) = -2 -2x
h®0