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Notes 2.1 - Limits
I. Limit Notation
A.) Two-sided Notation:
Read as “the limit of f(x) as x approaches a is L.”
lim ( )x a
f x L
B.) One-sided Notation:
Read as “the limit of f(x) as x approaches a from the right is L.”
Read as “the limit of f(x) as x approaches a from the left is L.”
lim ( )x a
f x L
lim ( )x a
f x L
II. Limit Definition
A.) Def: The function f has a limit L as x approaches c iff:
B.) Visual Representations:
lim ( ) lim ( ) lim ( )x c x c x c
f x L f x f x
2 11.) ( )
1
xf x
x
2 1, 1
2.) ( ) 11, 1
xx
g x xx
B.) Visual Reps(cont.)-
3.) ( ) 1h x x
III. Non-existent Limits
A.) fails to exist when:
1.) The right-side limit and left-side limit equal different real numbers.
2.) The are infinite oscillations .
3.) The limit(s) approach
lim ( )x a
f x
.
Ex. – Evaluate 0
1lim .x x
0
1lim x x
0
1lim x x
0
1lim Does Not Exist.x x
Although limits approaching infinity do not exist, we must still describe the behavior from both/each side(s)!!!
IV. Computational Techniques for Limits
A.) Properties of Limits: p. 61-62 of your text. KNOW THEM!!!
SUM, DIFFERENCE, PRODUCT, CONSTANT MULTIPLE, QUOTIENT, and POWER RULES
B.) Theorem – Polynomial and Rational Functions-
11 01.) If ( ) ... is any
polynomial fn and is any real number, then
n nn nf x a x a x a
c
11 0lim ( ) ... n n
n nx c
f x a c a c a
B.) Theorem – Polynomial and Rational Functions-
2.) If ( ) and ( ) are polynomial fns
and is any real number, then
f x g x
c
( ) ( )lim , if ( ) 0
( ) ( )x c
f x f cg c
g x g c
C.) Limit Properties and Theorems valid for one-sided limits as well.
V. Techniques for Evaluation
A.) Apply the theorems/properties (Substitute)
B.) Modify the expression:
1.) Simplify
2.) Factor
3.) Expand
4.) Multiply by conjugates
5.) Apply thm./prop. again
C.) Evaluate the following limits:
1.)
2.)
2
2
4lim
1x
x
x
22 4 00
2 1 3
2
2
4lim
2x
x
x
22 4 0???
2 2 0
2 2
2 2lim lim 2
2x x
x xx
x
2 2 4
3.)
4.)
2
2
4lim
2x
x
x
22 4 00
2 2 4
22
2lim
4x
x
x
2
2 2 4DNE
2 2 0
22
2 4lim
4 0x
x
x
22
2 4lim
4 0x
x
x
5.) 22
3lim
2 ( 5)x
x
x x
1DNE
0
22
22
3lim
2 ( 5)
3 1lim
02 ( 5)
x
x
x
x x
x
x x
6.) 4
4lim
2x
x
x
0???
0
4
4
4
2 24lim lim
lim 2 4 2 4
2 2
x
x x
x xx
x x
x
7.) 0
2 2lim x
x
x
0???
0
0
0 0
2 2 2 2lim
2 2
1lim lim
2 22 2
x
x x
x x
x x
x
xx x
1
2 2
8.) 2
4
3 49lim
4x
x
x
0???
0
4
4 4
3 7 3 7lim
44 10
lim lim 104
x
x x
x x
xx x
xx
14
9.) 3
4
64lim
4x
x
x
0???
0
2
4
2
4
4 4 16lim
4
lim 4 16
x
x
x x x
x
x x
48
VI. You Try!
1.)
2.)
3.)
4.)
3
2
2 16lim
2x
x
x
1
1lim
1x
x
x
0
9 3lim x
x
x
24
2
21
2lim
1x
x x
x
1
2
1
6
DNE
VII. Sandwich Theorem
GRAPHICALLY
A.) If ( ) ( ) ( ) for all in an open interval
containing the point (with the possible exception
at ) and lim ( ) lim ( ), then lim ( )x c x c x c
g x f x h x x
x c
x c g x L h x f x L
( )f x
( )h x
( )g x
B.) Example -
What do you know about the sin function?
2 2
0
1lim sinx
xx
11 sin 1
x
2 10 sin 1
x
2 2 2 210 sin 1x x x
x
2 2
0
1lim sin 0x
xx
2 2 210 sinx x
x 2 2 2
0 0 0
1lim 0 lim sin limx x x
x xx
2 2
0
10 lim sin 0
xx
x
C.) Example - 20
1lim cosx
xx
2
11 cos 1
x
2
11 cos 1x x x
x
2
1cosx x x
x
20 0 0
1lim lim cos limx x x
x x xx
20
10 lim cos 0
xx
x
20
1lim cos 0x
xx
VIII. Limit Theorems
0A.) lim cos 1
0B.) lim sin 0
0
sinC.) lim 1
0D.) lim 1
sin
IX. Proof of
Given the point P in the first quadrant on the unit circle:
The area of sector OAP = 21
2 2
0
sinlim 1
P (x, y)
A (1, 0)
O
(0, 0)
𝜃
1
Dropping a perp. From P to Q on the x-axis gives us triangle OPQ
The area of triangle OPQ = 1 1( ) sin cos
2 2x y
P (x, y)
A (1, 0)O
(0, 0)
𝜃1
Q (x,0)
Now, we can all agree that
Area Area Sect. OPQ OPA
1sin cos
2 2
P (x, y)
A (1, 0)O
(0, 0)
𝜃1
Q (x,0)
Extending OP to T and dropping a perp. from T to A on the x-axis gives us triangle OTA
The area of triangle OTA = 1 1
(1) tan tan2 2
P
A O
(0, 0)
1
Q
T(1, tan 𝜃)
𝜃
Now, we can all agree that
Area Area Sect. Area OPQ OPA OTA
1 1sin cos tan
2 2 2
P
A O
(0, 0)
1
Q
T(1, tan 𝜃)
𝜃
1 1sin cos tan
2 2 2
2 1 1sin cos tan
sin 2 2 2
1cos
sin cos
1 sincos
cos
sin 1cos
cos
0 0 0
sin 1lim cos lim lim
cos
0
sin1 lim 1
0
sinlim 1
X. Patching
In order to make our trigonometric limits look like A-D of II, we may need to “PATCH” the trig expression. After, we apply our limit properties and verify on our calculator.
A) Examples -
0
sin 31.) lim
x
x
x
0
sin 3lim x
x
x
0 0 0
sin 3 sin 3lim lim .lim
3
3 3
3x x x
x x
x xx x
x x
0 0
sinlim .lim 3
0 0
sin 3lim .lim 3
3x x
x
x 0LET 3 ; lim 0
xx
1 3 3
0
sin 32.) lim
5x
x
x
0
sin 3lim
5x
x
x
0 0 0
sin 3 sin 3 3 3 sin 3lim lim . lim
5 5 5
3
3 3 3x x x
x
x
x x x
x xx
3 31
5 5
0
sin 33.) lim
sin 2x
x
x
0
sin 3lim
sin 2x
x
x
0 0
sin 3 3 sin 3 2lim lim
sin
2
2 2 3 sin 2
3
3 2x x
x x
x x
x x x
x x x
3 31 1
2 2
3
0
sin4.) lim
x
x
x 2
0
sinlim sinx
xx
x 0 1 0
0
1 sin 3lim
cos3 sin 2x
x
x x
0
tan 35.) lim
sin 2x
x
x
0
sin 3lim
sin 2 cos3x
x
x x
0
1 sin 3 2 3lim
cos3 sin 23 2x
x x x
x xx x 3
2
0
1 sin 2 1lim
cos3 cos 2 7x
x
x x x
0
sec3 tan 26.) lim
7x
x x
x
0
1lim sec3 tan 2
7xx x
x
0
1 sin 2lim
cos3 cos 2 7x
x
x x x
0
1 sin 2 2 2lim
cos3 cos 2 7 72x
x x
x x xx