LIMITS

Name: _________________________________________________________

Mrs. Upham2019-2020

Lesson 1: Finding Limits Graphically and NumericallyWhen finding limits, you are finding the y-value for what the function is approaching. This can be done in three ways:

1. Make a table

2. Draw a graph3. Use algebra

Limits can fail to exist in three situations:1. The left-limit is

different than the right-side limit. y=|x|

x

2. Unbounded Behavior y= 1x2

3. Oscillating Behavior y=sin( 1x )

Verbally: If f(x) becomes arbitrarily close to a single number L as x approaches c from either side, then the limit of f(x) as x approaches c is L.Graphically: Analytically:

Numerically: From the table, limx→−5

f ( x )=3.4

x -5.01 -5.001 -5 -4.999 -4.99f(x) 3.396 3.399 3.4 3.398 3.3951. Use the graph of f(x) to the right to find

limx→−3

2x2+7 x+3x+3

2. Use the table below to find limx→2

g (x)

x 1.99 1.999 2 2.001 2.01f(x) 6.99 6.998 ERROR 7.001 7.01

3. Using the graph of H(x), which statement is not true?

a. limx→a−¿H (x)= lim

x →a+¿ H (x)¿¿¿

¿

b. limx→c

H ( x )=4

c. limx→b

H ( x ) does not exist

d. limx→c+¿H ( x )=2¿

¿

Lesson 2: Finding Limits AnalyticallyProperties of LimitsSome Basic LimitsLet b and c be real numbers and let n be a positive integer.

limx→c

f ( x )=f ( c ) limx→c

x=c limx→c

xn=cn

Methods to Analyze Limits1. Direct substitution2. Factor, cancellation technique3. The conjugate method, rationalize the numerator4. Use special trig limits of lim

x→0

sin xx

=1 or limx→0

1−cos xx

=0

Direct Substitution

1. limx→2

(3 x−5) 2. limx→4

3√x+4

3. limx→1

sin πx2

4. limx→7

x

5. If limx→c

f ( x )=7 then limx→c

5 f (x )

6. limx→c

√ f (x )

7. limx→c

[ f (x )]2

8. Given: limx→c

f ( x )=7 and limx→c

g ( x )=4, find:

a. limx→c

[ f ( x )+g (x)]

b. limx→c

f (g ( x ))

c. limx→c

g (f ( x ))

Limits of Polynomial and Rational Functions:

9. limx→0

x3+1x+1

10. limx→2

x3+1x+1

11. limx→−1

x3+1x+1

Limits of Functions Involving a Radical

12. limx→3

√ x+1−2x−3

Dividing out Technique

13. lim∆ x→

2 (x+∆x )−2x∆ x

14. Given f(x) = 3x + 2Find lim

h→ 0

f (x+h )−f (x)h

15.

limx→c

sin x

16. limx→c

cos x

17.limx→ π

2

sin x

18. limx→π

xcos x

Special Trigonometric Limits:

limx→0

sin xx

=1 limx→0

1−cos xx

=0

19. limx→0

tan xx 20. lim

x→0

sin 3 xx

The Squeeze TheoremIf h(x) < f(x) < g(x) for all x in an open interval containing c, except possibly at c itself, and if lim

x→ch (x )=L=lim

x→cg (x) then lim

x→cf (x ) exists and is equal to L.

4 – |x| < f(x) < 4 + |x|

Lesson 3: Continuity and One-Sided LimitsDefinition of ContinuityContinuity at a point:A function f is continuous at c if the following three conditions are met:

1. f(c) is defined2. lim

x→cf (x ) exists

3. limx→c

f ( x )=f (c )

Properties of continuity:Given functions f and g are continuous at x = c, then the following functions are also continuous at x = c.

1. Scalar multiple: b° f2. Sum or difference: f± g3. Product: f • g4. Quotient: fg , if g(c) ≠ 05. Compositions: If g is continuous at c and f is continuous at g©, then

the composite function is continuous at c, ( f ° g ) ( x )=f (g ( x ))

The existence of a Limit:

The existence of f(x) as x approaches c is L if and only if limx→c−¿ f ( x )=L ¿

¿ and lim

x→c+¿ f ( x )=L¿¿

Definition of Continuity on a Closed Interval:A function f is continuous on the closed interval [a, b] if it is continuous on the open interval (a, b) and lim

x→a+¿ f (x )=f (a)¿¿ and lim

x→b−¿ f ( x )= f (b )¿¿

Example 3: Given h ( x )={ −2x−5 ; x←23 ; x=−2

x3−6 x+3; x>−2 for what values of x is h not

continuous? Justify.

Example 4: If the function f is continuous and if f(x) = x2−4x+2

when x ≠ -2, then f(-2) = ?

Example 5: Which of the following functions are continuous for all real numbers x?

a. y = x23

b. y = ex

c. y = tan x

A) None B) I only C) II only D) I and IIIExample 6: For what value(s) of the constant c is the function g continuous over all the Reals? g ( x )={ cx+1 ; if x≤3c x2−1 ; if x>3

Lesson 4: The Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is an existence theorem which says that a continuous function on an interval cannot skip values. The IVT states that if these three conditions hold, then there is at least one number c in [a, b] so that f(c) = k.

1. f is continuous on the closed interval [a, b]2. f(a) ≠ f(b)3. k is any number between f(a) and f(b)

Example 1: Use the Intermediate Value Theorem to show that f(x) = x3 + 2x – 1 has a zero in the interval [0, 1].

Example 2: Apply the IVT, if possible, on [0, 5] so that f(c) = 1 for the function f(x) = x2+ x−1

Example 3: A car travels on a straight track. During the time interval 0 < t < 60 seconds, the car’s velocity v, measured in feet per second is a continuous function. The table below shows selected values of the function.t, in seconds 0 15 25 30 35 50 60v(t) in ft/sec -20 -30 -20 -14 -10 0 10

A. For 0 < t < 60, must there be a time t when v(t) = -5?B. Justify your answer.

Example 4: Find the value of c guaranteed by the Intermediate Value Theorem.

f(x) = x2 + 4x – 13 [0, 4] such that f(c) = 8

Lesson 5: Infinite LimitsDefinition of Vertical Asymptotes:

A vertical line x = a is a vertical asymptote if limx→a+¿ f (x )=±∞¿

¿ and/or limx→a−¿ f ( x )=±∞¿

¿

h ( x )= f (x )g (x)

has a vertical asymptote at x = c.

Properties of Infinite Limits:

Let c and L be real numbers and let f and g be functions such that limx→c

f ( x )=∞ and lim

x→cg ( x )=L

1. Sums or Difference: limx→c

[ f ( x )± g (x ) ]=∞

2. Product: limx→c

[ f ( x )g ( x ) ]=∞ , L>0

limx→c

[ f ( x )g ( x ) ]=−∞ , L<0

3. Quotient: limx→c

c g(x)f (x )

=0

Example 1: Evaluate by completing the table for limx→−3

1x2−9

x -3.5 -3.1 -3.01 -3.001

-3 -2.999 -2.99 -2.9 -2.5

f(x)

Example 2: Evaluate limx→1

1(x−1)2

Example 3: Evaluate limx→ 1+¿ x+1

x−1 ¿

¿

Example 4: Evaluate limx→1+¿ x2−3x

x−1 ¿

¿

Example 5: Evaluate limx→1+¿ x2

(x−1)2¿

¿

Example 6: Evaluate limx→0−¿( x2− 1x )¿

¿

Example 7: Evaluate limx→(−12 )

+¿ 6x2+ x−14x2−4 x−3

¿

¿

Example 8: Find any vertical asymptotes or removable discontinuities f ( x )= x−2

x2−x−2

Example 9: Determine whether the graph of the function has a vertical asymptote or a removable discontinuity at x = 1. Graph the function to confirm

f ( x )= sin( x+1)x+1