Calculus and Vectors – How to get an A+
1.5 Properties of Limits
A Limits Properties
We assume that lim f ( x) and lim g ( x) exist. Then:
Ex 1. Given lim f ( x) = −2 and lim g ( x) = 1 , use the limits
1. lim k = k
properties to find lim
x→a
x→a
x →3
x →3
2 f ( x) + g ( x)
− 4 g ( x)
x→3
x →a
2. lim x = a
x →a
3. lim[ f ( x) ± g ( x)] = lim f ( x) ± lim g ( x )
x →a
x→a
lim
x →a
x →a
x→3
7. lim[ f ( x)] = [ lim f ( x]
x →a
lim[−4 g ( x) ]
=
x→3
x→3
=
2 lim f ( x) + lim g ( x)
x→3
x→3
x→3
− 4 lim g ( x)
=
x→3
2(−2) + 1
− 4 +1 3
=
=
=
−4
4
−4 1
2 f ( x) + g ( x) 3
∴ lim
=
x→3 − 4 g ( x )
4
6. lim[ f ( x) g ( x)] = [ lim f ( x)][ lim g ( x)]
x →a
x→3
− 4 lim g ( x)
x →a
n
lim[2 f ( x) + g ( x )]
lim[2 f ( x)] + lim g ( x)
f ( x)
f ( x) xlim
5. lim
= →a
, lim g ( x) ≠ 0
x →a g ( x )
lim g ( x ) x→a
x →a
=
− 4 g ( x)
x→3
4. lim[cf ( x)] = c lim f ( x)
x →a
2 f ( x) + g ( x)
.
x →a
n
x→a
8. If P(x) is a polynomial function, then
lim P ( x) = P (a )
x →a
B Substitution
Substitution is the best strategy to find a limit.
Note: Substitution does not work if (by
substitution) you get one of the following 7
indeterminate cases:
0 ∞ ∞
∞ − ∞ 0×∞
1
∞ 0 00
0 ∞
Specific strategies are available to avoid an
indeterminate case to appear.
C Factoring
0
The indeterminate form may be eliminated by
0
factoring and canceling out the common factor that
generates zeros:
F ( x)
( x − a) f ( x)
f ( x)
lim
= lim
= lim
x →a G ( x )
x →a ( x − a ) g ( x )
x →a g ( x )
Note: Canceling out the common factor ( x − a ) is a
correct operation because lim means that
x→a
x2 − 2x +1
.
x →2
x −1
Ex 2. Compute lim
x 2 − 2 x + 1 2 2 − 2(2) + 1 4 − 4 + 1
=
=1
=
x →2
x −1
1
2 −1
x 2 − 2x + 1
=1
∴ lim
x→2
x −1
lim
x2 + x − 2
.
x→1
x −1
Let’s try substitution:
x 2 + x − 2 12 + 1 − 2 0
= (indeterminate case)
lim
=
x→1
x −1
1−1
0
First, we have to cancel out the common factor:
x2 + x − 2
( x − 1)( x + 2)
= lim( x + 2)
= lim
lim
x→1
x→1
x→1
x −1
x −1
Second, we can use substitution:
lim( x + 2) = 1 + 2 = 3
Ex 3. Compute lim
x→1
x approaches a but is not equal to a .
x2 + x − 2
=3
x→1
x −1
∴ lim
D Conjugate Radicals
0
you
0
may use the conjugate radicals to cancel out the
common factor that generates zeros.
When dealing with the indeterminate form
Ex 4. Compute lim
x→0
x+4 −2
.
x
0
).
0
( x + 4) − 2 2
Substitution does not work (case
x+4 −2
x
x+4 +2
= lim
=
x( x + 4 + 2)
x
1
1
1
= lim
= lim
=
=
x →0 x ( x + 4 + 2 )
x→0 x + 4 + 2
0+4 +2 4
lim
x→0
x+4 +2
1.5 Properties of Limits
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2
x →0
Calculus and Vectors – How to get an A+
E Change of Variables
x −1
.
By changing the variable, the process of canceling Ex 5. Change the variable to compute lim 3
x→1 x − 1
the common factor may be simplified.
1
1
Note. If the change of variable is u = g (x) then:
as x → a , u → g (a)
x = x2,
2
u =
1
(x 6 )2
3
=
1
x3
x →1 ⇒ u
x −1
lim
x→1 3
x −1
= lim
u →1
F Absolute Value Function
When dealing with an absolute value function,
rewrite it as piece-wise defined function according
to:
⎧ f ( x), f ( x) ≥ 0
| f ( x) |= ⎨
⎩− f ( x ), f ( x ) < 0
b) lim x | x |
u −1
2
=
1
x2
= x
=1
3
u →1 u 2
1
( x 6 )3
−1
(u − 1)(u 2 + u + 1)
=
u →1
(u − 1)(u + 1)
= lim
2
u + u +1 1 +1+1 3
x −1 3
=
= , ∴ lim
=
3
1
x
→
u +1
1+1
2
x −1 2
Ex 6. Compute each limit. Draw a diagram to illustrate.
|x|
a) lim
x→0 x
⎧ x, x ≥ 0
| x | ⎧1, x > 0
| x |= ⎨
⇒
=⎨
x ⎩− 1, x < 0
⎩− x , x < 0
| x|
= lim− (−1) = −1,
x→0 x
x →0
| x|
∴ lim
= DNE
x→0 x
lim−
lim+
x →0
|x|
= lim 1 = 1
x x →0 +
x2
x→0 | x |
⎧⎪ x 2 , x ≥ 0
x | x |= ⎨
⎪⎩− x 2 , x < 0
lim x | x |= lim− (− x 2 ) = 0,
x →0
∴ lim x | x |= 0
x →0
= x, u =
1
→ 16
= lim
3
3
c) lim
x→0
x →0 −
1
x = x 3 , LCD = 6, u = x 6
lim x | x |= lim+ x 2 = 0
x →0 +
x →0
x 2 ⎧ x, x > 0
=⎨
| x | ⎩− x , x < 0
lim−
x→0
x2
= lim x = 0,
| x | x →0 −
lim+
x→0
2
x
=0
x→0 | x |
∴ lim
Reading: Nelson Textbook, Pages 40-45
Homework: Nelson Textbook: Page 45 #4f, 7, 8, 9, 10, 13b, 14b, 16, 17
1.5 Properties of Limits
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2
x2
= lim (− x) = 0
| x | x →0 +