Aim: What are some techniques for evaluating limits?

Aim: Evaluating Limits Course: Calculus

Do Now:

Aim: What are some techniques for evaluating limits?

Sketch

f (x) 3x 1

x 2


Do Now:

Aim: What are some techniques for evaluating limits?

If 1 1

( ) 3lim 2, find lim ( )

1x x

f xf x

x


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


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.


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


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


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


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


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


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


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


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.


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=


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


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


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

Aim: What are some techniques for evaluating limits?

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