**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 - x

^{2}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 - x

^{2}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 - x

^{2}

lim ([4 - 2(x + h) - (x+h)

^{2}] - [4 - 2x - x

^{2}])/h =

^{h®0}

lim ([4 - 2x -2h -(x

^{2}+ 2xh + h

^{2}] - 4 + 2x + x

^{2})/h =

^{h®0}

lim (4 - 2x -2h -x

^{2}-2xh - h

^{2}- 4 + 2x + x

^{2})/h =

^{h®0}

lim (- 2h - 2xh - h

^{2})/h =

^{h®0}

lim (- 2 -2x -h) = -2 -2x

^{h®0}