aim: what are some techniques for evaluating limits?
DESCRIPTION
Aim: What are some techniques for evaluating limits?. Do Now:. Sketch. Aim: What are some techniques for evaluating limits?. If. Do Now:. y -int. –. x -int. –. Vertical asymptotes –. Horizontal asymptotes –. y = 3. Plot several points. x = 2. Do Now. Graph:. f (0). - PowerPoint PPT PresentationTRANSCRIPT
Aim: Evaluating Limits Course: Calculus
Do Now:
Aim: What are some techniques for evaluating limits?
Sketch
f (x) 3x 1
x 2
Aim: Evaluating Limits Course: Calculus
Do Now:
Aim: What are some techniques for evaluating limits?
If 1 1
( ) 3lim 2, find lim ( )
1x x
f xf x
x
Aim: Evaluating Limits Course: Calculus
10
5
-5
-10
-10 10
10
5
-5
-10
-10 10
f2 x = 3x-1
x-2
Do Now
Graph:
f (x) 3x 1
x 2
y = 3
y-int. –
x-int. –
Vertical asymptotes –
Horizontal asymptotes –
Plot several points
x = 2
f(0) = 1/2 (0, 1/2)
x – 2 = 0, x = 2
x -2 -1 1 4 5
y 7/4 4/3 -2 11/2 14/3
1/3
If degree of p = degree of q, then the line y = an/bm is a horizontal asymptote.
q(x) = 0
y = 3
Aim: Evaluating Limits Course: Calculus
Asymptotes of Rational Functions
f (x) p(x)
q(x)
anxn an 1xn 1 ....a1x1 a0
bmxm bm 1xm 1 ....b1x1 b0
,
Let f be the rational function given by
where p(x) and q(x) have no common factors1. The graph of f has vertical asymptotes at the zeros of q(x).2. The graph of f has at most one
horizontal asymptote, as follows: a) If degree of p < degree of q, then the x-axis (y = 0) is a horizontal asymptote. b) If degree of p = degree of q, then the line y = an/bm is a horizontal asymptote. c) If degree of p > degree of q, then the graph of f has no horizontal asymptote.
Aim: Evaluating Limits Course: Calculus
1. If p is a polynomial function and c is a real number, then
2. If r is a rational function given by r(x) = p(x)/q(x), and c is a real number such that q(c) 0, then
direct substitution
Limits of Polynomial and Rational Functions
)()(lim cpxpcx
0)(,)(
)()()(lim
cq
cq
cpcrxr
cx
)6(lim 2
1
xx
x
1lim 3 2x
x
6)1(12
6
2
1
6lim 2 3
3x
x x
x
4
2
-2
-4
-6
5
f x = x2+x -6
x+3
Aim: Evaluating Limits Course: Calculus
Limits of Composite Functions
If f and g are functions such that
lim ( ) and lim ( ) ( ),x c x L
g x L f x f L
then lim ( ) ( )x c
f g x f L
2
0lim 4x
x
limx
x
= 4 = 2
2
0lim 4x
x
g(x) f(x)
f(g(x))
= 2
4
Aim: Evaluating Limits Course: Calculus
Limits of Trigonometric Functions
Let c be a real number in the domain of the given trig function1. limsin sin
x cx c
2. limcos cosx c
x c
3. limtan tanx c
x c
4. limcot cotx c
x c
5. limsec secx c
x c
6. limcsc cscx c
x c
lim( cos )x
x x lim limcos
x xx x
cos 1
2
0limsinx
x
2
0lim sinx
x
20 0
limx c
x c
Aim: Evaluating Limits Course: Calculus
Evaluating Limits as x a Finite Number c
To evaluate a limit algebraically as x approaches a finite number c, substitute c into the expression.
1. If the answer is a finite number, that number is the value of the limit.
2. If the answer is of the form 0/0, we have an indeterminate form. •Factor the numerator or denominator, simplify, substitute for c
•Rationalize the numerator or the denominator, simplify, substitute•Simplify complex fraction, substitute
Aim: Evaluating Limits Course: Calculus
Divide Out
Dividing Out/Factoring Technique
Find the 3
6lim
2
3
x
xxx
Problem: direct substitution results in an indeterminate form.
-2
-4
-6
-5
u x = x2+x -6
x+3
3
6lim
2
3
x
xxx 3
)3)(2(lim
3
x
xxx
52)3()2(lim3
xx
Note: this technique works only when direct substitution results in zeros in both numerator and denominator.
q(-3) = 0
0
0
Aim: Evaluating Limits Course: Calculus
Direct substitution results in an 0 in both numerator and denominator and will not yield a limit.
Factor Out
Dividing Out/Factoring Technique
Find the 1
1lim 231
xxx
xx
1
1lim 231
xxx
xx )1)(1(
1lim 21
xx
xx
)1(
1lim 21
xx
2
1
)11(
12
1.5
1
0.5
-0.5
-1 1 2
v x = x-1
x3-x2+x -1
Aim: Evaluating Limits Course: Calculus
Rationalize numerator
x
x 11
Rationalizing Technique
Find the x
xx
11lim
0
Direct substitution results in an 0 in both numerator and denominator and will not yield a limit.
11
1111
x
x
x
x
( 1) 1
1
x
x x x
1
x
x x x
11
1
x
x
xx
11lim
0
2
1
11
1
11
1lim
0
xx
Aim: Evaluating Limits Course: Calculus
Using Technology Technique
Table of values starting at –0.003
Graph
4
2
w x = 1+x 1
x
Take average: (2.7196 + 2.7169)/2 2.71825
Find the x
xx
1
)1(lim0
= e
Use zoom and trace to find coordinates that are equidistant from x = 0 and take average of corresponding y’s.
Aim: Evaluating Limits Course: Calculus
Squeeze Theorem
If h(x) < f(x) < g(x) for all x in an open interval containing, c, except possibly at c itself, and if
lim ( ) lim ( )x c x c
h x L g x
then exists and is equal to L. lim ( )x c
f x
2 sin1 1
2
x x
x Given:
0
sinFind lim
x
x
x
2
0lim 1
2x
x
0lim1x= 0 = 0
0
sinlim 0x
x
x
h(x) g(x)
f(x)
L=
Aim: Evaluating Limits Course: Calculus
Special Trig Limits
0
tanlimx
x
x
0
sinlim 1x
x
x
0
1 coslim 0x
x
x
0
1lim sin
cosxx
x x
sintan
cos
xx
x
0
sincos
limx
xx
x
0
sin 1lim
cosx
x
x x
0 0
sin 1since lim 1 and lim 1
cosx x
x
x x
0
tanlim 1 1 1x
x
x
[ ( ) ( )]limx c
f x g x LK
x is in radians0
cos 1lim 0x
x
x
Aim: Evaluating Limits Course: Calculus
More Special Trig Limits
0
sinlimx
ax a
bx b
0lim 1
sinx
x
x
0lim
sinx
ax a
bx b
0
sinlim
sinx
ax a
bx b
0lim 1
tanx
x
x
0
sin4limx
x
x 0
sin44 lim
4x
x
x
let y = 4x0
sin4 lim
y
y
y
4 1 4 0
sinlim 1x
x
x
0
4sin4lim
4x
x
x
multiply top and bottom by 4
Aim: Evaluating Limits Course: Calculus
Model Problems
Evaluate:
0
sin7limx
x
x
0
sin7lim
7x
x
x
0
sinlim
7x
x
x
2
0
1 coslimx
x
x
2
0
coslimx
x
x
0
sin3lim
sin8x
x
x
0
4lim
tanx
x
x
2
20lim
1 cosx
x
x