1.2 Evaluating Limits Algebraically
Properties of Limits:
lim f ( x )
x→c
7
1. lim b = b
ex : lim = 7 lim = 7 lim 3 = 3
x →5
x →c
x →3
f ( x) = 7
x→2
x
f(x)
7
7
7
7
7
7
5
7
f ( x) = x
2. lim x = c
x →c
ex: x
4.9 4.99 4.999 4.9999 5.0001 5.001 5.01
1.9 1.99
1.999 2.001 2.01
f(x) 1.9 1.99
1.999 2.001 2.01
3. lim x n = c n (when n is a positive integer)
x →c
f ( x ) = x2
Ex:
f ( x ) = x3
lim x 2 = 22 = 4
lim x 3 = 23 = 8
x→2
x →2
4. Scalar Multiple Property
lim b ⋅ f ( x ) = b ⋅ lim f ( x ) *The constant factor “b” can move through limits. This works for any constant “b” and any
x →c
x →c
function.
f ( x ) = 3x 2
lim 3 x 2 = 12
x→2
2
2
lim 3 x = 3 ⋅ lim x = 3 ⋅ 4 = 12
x→2
x →2
5. Sum or Difference Property
lim  f ( x ) ± g ( x )  = lim f ( x ) ± lim g ( x )
x →c
x →c
x →c
ex:
f ( x ) = x2
2
y = x +2
g ( x) = x
lim  f ( x ) + g ( x )  = lim  x 2 + x 
x→ 2
x→2
lim  x + x  = lim x + lim x = 4 + 2 = 6
x→2
x→2
x→ 2
2
2
6. Product Rule
lim  f ( x ) ⋅ g ( x )  = lim f ( x ) ⋅ lim g ( x ) *The limit of a product is the product of the limits.
x →c
x →c
x →c
7. Quotient Rule
f ( x)
 f ( x )  lim
x →c
lim 
=
* The limit of a quotient is the quotient of the limits.

x →c g ( x )
lim
g
x
(
)

 x →c
n
8. lim  f ( x )  = lim f ( x ) 
x →c
 x →c

n
Examples:
lim  x 2 − 2 x + 5
x →3
Property (5) =
lim x 2 − lim 2 x + lim 5
x →3
x →3
x →3
Property (4) =
lim x 2 − 2 lim x + lim 5
x →3
x →3
x →3
Property (3) =
32 − 2 ⋅ 3 + 5 = 9 − 6 + 5 = 3 + 5 = 8
☼ You can use direct substitution to calculate the limit of a function IF the function is continuous at “c”.
* In the example you will just substitute in the “c” value into the equation.
2
lim ( 3) − 2 ( 3) + 5 = 8

x→3 
☼ A function is not continuous when there are holes at “c”. If you try direct substitution it will yield
0
whenever there is a hole at “c”
0
* This is most important to keep in mind when working with piecewise defined functions, where a break or jump in the
function may exist.
* For a piecewise defined function to have a limit the function’s pieces have to get the same number when you do direct
substitution.
☼ Also when you do direct substitution and get
#
not zero, it is not the answer to the limit, but it does tell you where an asymptote is,
0
and that a limit does not exist at that point for that function.
☼ Direct substitution will work for any polynomial, as well as any trig function (where it is defined) and any log function (on its
domain).
To calculate limits analytically there are three main approaches.
First try direct substitution, keeping in mind what your function looks like.
If you get
0
#
, then try factor and cancel second. (If you get then the limit does not exist (DNE))
0
0
**** Remember that piecewise functions are the exception to direct substitution; may be able to get an answer, but that answer may
not be the answer to the limit.
Example:
x2 − 5x + 6
0
Find : lim
Direct substitution yields . Now try to factor and cancle.
x →2
x−2
0
= lim
x→2
= lim
x→2
X 2 − 5X + 6
X −2
( X − 3) ( X − 2 )
( X −2)
= lim ( X − 3)
x→2
Now use direct substitution!
= 2 − 3 = −1
Factor and Cancel:
x3 − 6 x 2 − 4 x + 24
1. lim
We can use synthetic division to help us factor here! −2 1 − 6 − 4 24
x →−2
x+2
− 2 16 − 24
1
( x + 2 ) ( x − 8 x + 12 )
( x+2)
-8
12
0
2
= lim
x →−2
2
= ( −2 ) − 8 ( −2 ) + 12 = 32 * Canceling (x+2) also shows us that if we used direct substitution
at the start, with (-2), we would have gotten
0
! Now that we have factored and canceled we can use direct substitution to find the
0
answer to the limit.
We could have also factored this problem from the start…it just takes a little more work. Compare for yourself:
x3 − 6 x 2 − 4 x + 24
x →−2
x+2
2
x ( x − 6) − 4 ( x − 6)
= lim
x →−2
x+2
( x − 6) ( x2 − 2)
= lim
x →−2
x+2
1. lim
= lim
( x − 6) ( x + 2 ) ( x − 2)
x+2
x →−2
Now we will use direct substitution
= lim ( x − 6 )( x − 2 ) = ( −2 − 6 )( −2 − 2 ) = ( −8 )( −4 ) = 32 with a hole at (-2,32)
x →−2
Either way you choose to do this problem is correct; it just might not always be obvious how to factor a problem such as this. In the
end, for either method the answer is still 32.
**** Don’t forget to write down = lim all the way through your work, until you do direct substitution!
x→−2
4
2
x + 3x + x − 105
0
* First try direct substitution! Here we get , so we will need to try something else.
3
x →−3
x − 4 x + 15
0
2. lim
We will try factoring with the help of synthetic division next.
Using our “c” value will help us with factoring.
−3 1
0
3
1
− 105
− 3 9 − 36
1 − 3 12 − 35
105
0
0
0
−3 1 0 − 4 15 *Getting 0s helps us find our indetermanent!
−3
5 0
0
= 0 Limit answer would be 0!
5
5
*If we would have gotten = Does Not Exist (DNE) (asymptote here instead)
0
*If we would have gotten
Now we will see what happens when we factor and cancel:
( x + 3 ) ( x − 3x + 12 x − 35)
= lim
( x + 3 ) ( x − 3x + 5 )
3
x →−3
2
2
Try direct substitution again…
-3 is no longer in our list of p/q so 0/0 will not happen again.
−3 1 − 3 12
35
− 3 18
1 − 6 30
− 90
− 125
−3 1 − 3
5
=
− 3 18
1 − 6 23
−125
23
************************************************************************************************************
The third method for finding limits is conjugation!
3. lim
x −5
x − 25
= lim
x − 25
x −5  x +5
⋅ 
= lim
 = xlim
x − 25  x + 5  →25 x − 25
x + 5 x →25
x → 25
x → 25
(D.S. yields
0
!)
0
(
)
)(
(
)
1
(
x +5
)
=
1
1
1
=
=
25 + 5 5 + 5 10
* You cannot use synthetic division, like in the previous examples, with radicals; although, this problem could have also been factored
from the start.
3. lim
x → 25
x −5
= lim
x − 25 x →25
x −5
( x)
2
− ( 5)
2
= lim
x → 25
x −5
(
x +5
)(
x −5
)
1
1
1
1
=
=
=
x +5
25 + 5 5 + 5 10
= lim
x → 25
FIND:
x + 1 −1
0
(D.S. yields !)
x
0
x + 1 −1 x + 1 +1
x + 1−1
= lim
⋅
= lim
= lim
x →0
x
x + 1 + 1 x → 0 x x + 1 + 1 x →0 x
4. lim
x →0
(
=
)
x
(
)
x +1 +1
= lim
x →0
1
=
x +1+1
1
1
1
1
=
=
=
0 +1 +1
1 +1 1+1 2
x+7
0
(D.S. yields !)
x →−7 3 − 2 − x
0
5. lim
(
)
(
)
( x + 7) 3 + 2 − x
( x + 7) 3 + 2 − x
x+7
3+ 2− x
⋅
= lim
= lim
=
x →−7 3 − 2 − x 3 + 2 − x
x →−7
x →−7
9 − (2 − x)
9−2+ x
= lim
= lim
x →−7
( x + 7 ) (3 +
x+7
2− x
) = 3+
2 − ( −7 ) = 3 + 9 = 3 + 3 = 6
1
1
−
0
6. lim x + 4 4 (D.S. yields !) *We have to simplify the complex fraction first!
x →0
x
0
1
1
1
1
−
− 4 x+4
(
) = lim 4 − ( x + 4 ) = lim 4 − x − 4 = lim − x =
= lim x + 4 4 = lim x + 4 4 ⋅
x →0
x →0
x
x
4 ( x + 4 ) x → 0 4 x ( x + 4 ) x → 0 4 x ( x + 4 ) x →0 4 x ( x + 4 )
−1
−1
=
x →0 4 ( x + 4 )
16
= lim
The Squeeze Theorem
☼ Let
f , g , and h be functions satisfying h ( x ) ≤ g ( x ) ≤ f ( x ) for all X in some open interval containing C, with the possible
exception that the inequalities do not hold at C.
If
f and h have the same limit as XC, say lim f ( x ) = 2 and lim h ( x ) = L then g also has the limit as XC.
x →c
**Example** lim g ( x ) =
x →c
x →c
L , lim h ( x ) = L , and lim f ( x ) = L
x →c
x →c
http://commons.wikimedia.org/wiki/File:Generic_Squeeze_or_Sandwich_Theorem_Representation.svg
*Y values of h are higher than f for some open interval where
have the same limit as the red and green functions.
h and f are above and below g forcing the blue to
☼Since there is an open interval containing C where
the lim =
x →c
h ( x ) ≤ g ( x ) ≤ f ( x ) then by the squeeze theorem
g ( x) = L .
****************************************************************************************************
****************************************************************************************************
Find: lim
sin θ
θ →0
Let
Remember that A =
θ
1 2
rθ
2
θ be an angle in radian measure where 0 < θ <
π
2
h
1
sin θ h
=
cos θ 1
≥
≥
tan θ 1 2
1
≥ (1) θ ≥ ⋅1 ⋅ sin θ
2
2
2
**Property of inequality:
a>b→
To solve:
1) Multiply by
2
1
θ
to get
≥
≥1
sin θ
cos θ sin θ
1 1
<
a b
sin θ
2) Take the Reciprocal to get: cos θ ≤
π
2
≤1
π
*True when 0 < θ <
Also true when −
θ
2
** cos ( −θ ) ≤
<θ < 0
sin ( −θ )
−θ
− sin θ
cos θ ≤
≤1
−θ
3) So we have found an open interval,
≤1
sin θ
 π π
is trapped in between cos θ and 1.
 − ,  where
θ
 2 2
☼ So by the squeeze theorem you can see that
lim cos θ = 1 and lim1 = 1 so… lim
θ →0
θ →0
θ →0
sin θ
θ
=1
*Example 2:
lim
1 − cos θ
θ
θ →0
=0
Proof:
lim
θ →0
1 − cos θ
θ
lim
θ →0
sin θ ⋅ sin θ
1 − cos θ 1 + cos θ
1 − cos 2 θ
sin 2 θ
⋅
= lim
= lim
= lim
=
θ
→
0
θ →0
θ
→
0
θ
→
0
θ
θ (1 + cos θ )
θ (1 + cos θ )
θ (1 + cos θ )
1 + cos θ
= lim
sin θ
θ
⋅
sin θ
sin θ
sin θ
0
= lim
⋅ lim
= 1⋅ = 0
(1 + cos θ ) θ →0 θ θ →0 (1 + cos θ ) 2