LIMITS
Name: _________________________________________________________
Mrs. Upham
2019-2020
Lesson 1: Finding Limits Graphically and Numerically
When 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 graph
3. Use algebra
Limits can fail to exist in three situations:
1. The left-limit is
different than
the right-side
limit.
π¦ = |π₯|
π₯
2. Unbounded
Behavior
π¦ = 1
π₯2
3. Oscillating
Behavior
π¦ = π ππ (1
π₯)
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, limπ₯ββ5
π(π₯) = 3.4
x -5.01 -5.001 -5 -4.999 -4.99
f(x) 3.396 3.399 3.4 3.398 3.395
1. Use the graph of f(x) to the right to find
limπ₯ββ3
2π₯2 + 7π₯+3
π₯+3
2. Use the table below to find limπ₯β2
π(π₯)
x 1.99 1.999 2 2.001 2.01
f(x) 6.99 6.998 ERROR 7.001 7.01
3. Using the graph of H(x), which statement is not true?
a. limπ₯βπβ
π»(π₯) = limπ₯βπ+
π»(π₯)
b. limπ₯βπ
π»(π₯) = 4
c. limπ₯βπ
π»(π₯) does not exist
d. limπ₯βπ+
π»(π₯) = 2
Lesson 2: Finding Limits Analytically
Properties of Limits
Some Basic Limits
Let b and c be real numbers and let n be a positive integer.
limπ₯βπ
π(π₯) = π(π)
limπ₯βπ
π₯ = π limπ₯βπ
π₯π = ππ
Methods to Analyze Limits
1. Direct substitution
2. Factor, cancellation technique
3. The conjugate method, rationalize the numerator
4. Use special trig limits of limπ₯β0
sin π₯
π₯= 1 or lim
π₯β0
1βcos π₯
π₯= 0
Direct Substitution
1. limπ₯β2
(3π₯ β 5)
2. limπ₯β4
βπ₯ + 43
3. limπ₯β1
sinππ₯
2
4. limπ₯β7
π₯
5. If limπ₯βπ
π(π₯) = 7 then limπ₯βπ
5π(π₯)
6. limπ₯βπ
βπ(π₯)
7. limπ₯βπ
[π(π₯)]2
8. Given: limπ₯βπ
π(π₯) = 7 and limπ₯βπ
π(π₯) = 4, find:
a. limπ₯βπ
[π(π₯) + π(π₯)]
b. limπ₯βπ
π(π(π₯))
c. limπ₯βπ
π(π(π₯))
Limits of Polynomial and Rational Functions:
9. limπ₯β0
π₯3+1
π₯+1
10. limπ₯β2
π₯3+1
π₯+1
11. limπ₯ββ1
π₯3+1
π₯+1
Limits of Functions Involving a Radical
12. limπ₯β3
βπ₯+1β2
π₯β3
Dividing out Technique
13. limβπ₯β
2(π₯+ βπ₯)β2π₯
βπ₯
14. Given f(x) = 3x + 2
Find limββ0
π(π₯+β)βπ(π₯)
β
15. limπ₯βπ
sin π₯
16. limπ₯βπ
cos π₯
17. limπ₯β
π
2
sin π₯
18. limπ₯βπ
π₯ cos π₯
19. limπ₯β0
tan π₯
π₯
20. limπ₯β0
sin 3π₯
π₯
The 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 limπ₯βπ
β(π₯) = πΏ = limπ₯βπ
π(π₯) then limπ₯βπ
π(π₯) exists and is equal to L.
4 β |π₯| < f(x) < 4 + |π₯|
Special Trigonometric Limits:
limπ₯β0
sin π₯
π₯= 1 lim
π₯β0
1βcos π₯
π₯= 0
Lesson 3: Continuity and One-Sided Limits
Definition of Continuity
Continuity at a point:
A function f is continuous at c if the following three conditions are met:
1. f(c) is defined
2. limπ₯βπ
π(π₯) exists
3. limπ₯βπ
π(π₯) = π(π)
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: π Β° π
2. Sum or difference: fΒ± g
3. Product: f β’ g
4. Quotient: π
π , if g(c) β 0
5. Compositions: If g is continuous at c and f is continuous at gΒ©, then the
composite function is continuous at c, (π Β° π)(π₯) = π(π(π₯))
The existence of a Limit:
The existence of f(x) as x approaches c is L if and only if limπ₯βπβ
π(π₯) = πΏ and
limπ₯βπ+
π(π₯) = πΏ
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 limπ₯βπ+
π(π₯) = π(π) and limπ₯βπβ
π(π₯) = π(π)
Example 3: Given β(π₯) = {β2π₯ β 5 ; π₯ < β2
3 ; π₯ = β2
π₯3 β 6π₯ + 3 ; π₯ > β2 for what values of x is h not
continuous? Justify.
Example 4: If the function f is continuous and if f(x) = π₯2β4
π₯+2 when x β -2, then
f(-2) = ?
Example 5: Which of the following functions are continuous for all real numbers x?
a. y = π₯2
3
b. y = ex
c. y = tan x
A) None B) I only C) II only D) I and III
Example 6: For what value(s) of the constant c is the function g continuous over all
the Reals? π(π₯) = {ππ₯ + 1 ; ππ π₯ β€ 3
ππ₯2 β 1 ; ππ π₯ > 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) β 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) = π₯3 + 2x β 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) = π₯2 + π₯ β 1
Example 3: A car travels on a straight track. During the time interval 0 < t < 60
seconds, the carβs 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 60
v(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 β 13 [0, 4] such that f(c) = 8
Lesson 5: Infinite Limits
Definition of Vertical Asymptotes:
A vertical line x = a is a vertical asymptote if limπ₯βπ+
π(π₯) = Β±β and/or
limπ₯βπβ
π(π₯) = Β±β β(π₯) = π(π₯)
π(π₯) 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 limπ₯βπ
π(π₯) = β and
limπ₯βπ
π(π₯) = πΏ
1. Sums or Difference: limπ₯βπ
[π(π₯) Β± π(π₯)] = β
2. Product: limπ₯βπ
[π(π₯)π(π₯)] = β , πΏ > 0
limπ₯βπ
[π(π₯)π(π₯)] = ββ , πΏ < 0
3. Quotient: limπ₯βπ
ππ(π₯)
π(π₯)= 0
Example 1: Evaluate by completing the table for limπ₯ββ3
1
π₯2β9
x -3.5 -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9 -2.5
f(x)
Example 2: Evaluate limπ₯β1
1
(π₯β1)2
Example 3: Evaluate limπ₯β1+
π₯+1
π₯β1
Example 4: Evaluate limπ₯β1+
π₯2β3π₯
π₯β1
Example 5: Evaluate limπ₯β1+
π₯2
(π₯β1)2
Example 6: Evaluate limπ₯β0β
(π₯2 β 1
π₯)
Example 7: Evaluate limπ₯β(
β1
2)
+
6π₯2+π₯β1
4π₯2β4π₯β3
Example 8: Find any vertical asymptotes or removable discontinuities π(π₯) = π₯β2
π₯2βπ₯β2
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
π(π₯) = sin(π₯ + 1)
π₯ + 1
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