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Calculus 12
Evaluating Limits Analytically
3 techniques:
1. Direct Substitution sections A - D
2. Dividing Out sections A - D
3. Rationalizing the numerator sections E
A. Evaluating limits by direct substitution.
1.
3.
5.
lim
x 8
2.
x
lim
x 
( x 2  7)
lim
x 2
lim
x  4
4.
225
 x 3  3x  2
x  3
2 x
lim
[( x  4)( x  1)]
B. Evaluate each of the following limits by factoring first.
1.
lim 2r 2  6r
r 3 r 3
3.
x4
(yes, you can factor
x 4 x 2
2.
m 2  5m  4
m  1 m 3  1
lim
lim
- strain your brain a little)
lim x 2  81
4.
(don’t rationalize, factor)
x  9 3 x
5.
lim x 3  32
(you can factor
x 2 x2
--)
6. Explain why it is not completely correct to say
but it is correct to say that
x 2  64
 ( x  8) ,
x 8
lim x 2  64
lim

( x  8)
x 8 x 8
x 8
C. Determine the coordinates of any holes in the graphs of the functions below.
x 2  4x
2. f ( x)  2
2 x  8x
2x  4
1. f ( x) 
x2
3. h( x) 
x3  4x
(there are two)
x 2  2x
m( x) 
4.
x 3
x 9
x4 1
5. h( x) 
x 1
D. Evaluate each of the following limits. Begin by simplifying.
You may then have to factor.
1.
lim ( x  1) 2  25
x4
x4
2.
1
1

x3 3
4.
x  0 2x
1
1
b
3.
x  1 b  1
lim
5.
lim (2  h) 2  4
h0
h
lim
lim 2( x  h) 2  2( x  h) 2
h0
h
Answer Key
2
5
2. 225
3.  2  7
4.
B. 1. 6
2. 3
3. 4
4. 108
6.
omits the fact that x = 8 is a value that could be included giving an error in
A.1.
8
function,
5.  2  3 2
5. 80
is only interested in the values approaching 8
C. 1. (2,2)
2. (0,
1
1
) (4, )
2
2
3. (0,-2) (-2,-4)
5.
; (1, 4)
6.
; (0,1) (1,4)
D. 1.
2.
3.
4.
5.
4. (9,
1
)
6
E. Evaluating limits by rationalizing an irrational numerator or denominator.
1.
x 1
x  1 x 1
lim
Answer Key to E
2.
1 x 1
x0
x
lim
x  10
3.
x  10 x  10
lim
4.
5 x
x  25 25  x
lim
1.
1
2
2. -
1
2
3.
1
2 10
4.
5.
9 x
x 9 x 3
1
10
lim
5. -6