AP Calc Notes: L-4 Limit Properties/Limits by Substitution
Properties of Limits
Suppose k is a constant and lim f ( x ) and lim g ( x ) both exist. Then the following properties hold:
x →a
x →a
1. lim k = k
k is a constant, no matter what x does, k = k.
2. lim x = a
As x gets closer to a, x gets closer to a.
x →a
x →a
3. lim kf ( x ) = k lim f(x)
x →a
x→a
A constant (not a variable) may be factored out of the limit.
4. lim  f ( x ) + g ( x )  = lim f(x) + lim g(x)
x →a
x→a
x→a
5. lim  f ( x ) g ( x )  = lim f(x) lim g(x)
x →a
x→a
x →a
6. lim
x →a
The limit of a sum is the sum of the limits.
The limit of a product is the product of the limits.
f(x)
f ( x ) lim
= x→ a
( lim g(x) ≠ 0) The limit of a quotient is the quotient of the limits.
g ( x ) lim g(x) x→ a
x→a
7. If f is continuous at lim g(x) , then lim f ( g ( x ) ) = f( lim g(x) ).
x→a
x →a
x→a
Ex: If lim f ( x ) = 12 , lim g ( x ) = −3 and lim f ( x ) = f (−3) = 5 , evaluate
x→2
x→2
a. lim 7 =
x→2
b. lim x =
x→2
c. lim 5 f ( x ) =
x→2
d. lim  f ( x ) + g ( x )  =
x→2
x →−3
e. lim
x→2
f ( x)
=
g ( x)
f. lim  g ( x )  =
x→2
4
g. lim
x→2
f ( x) =
h. lim f ( g ( x ) ) =
x→2
Evaluating Limits by Direct Substitution
Ex: Use the properties of limits to evaluate lim ( x 2 − 4 x + 5 )
x →3
lim ( x 2 − 4 x + 5 ) =
x →3
Many (not all!) limits can be evaluated by “direct substitution.”
Ex: lim e− x cos x =
x →0
Ex: lim
x2 − 5
=
x+3
Ex: lim
x2 − 4
=
x−4
x →3
x→4
When doesn’t it work?
I. Quotients
3 cases for quotients:
0
a
a
0
Ex: lim
x→2
0
0
x2 − 4
x−2
x
II. Piece-wise defined functions
Ex: lim [ x ] where [x] is the greatest integer function:
x→2
[x] = the greatest integer ≤ x.
x
 x2
x ≤1

Ex: lim f ( x ) and lim f ( x ) where f ( x )  x 1 < x ≤ 4
x →1
x→4
5 − x
x>4

x