Download Calculus 12

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
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
Related documents