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.

𝑦 = |𝑥|

𝑥

2. Unbounded

Behavior

𝑦 = 1

𝑥2

3. Oscillating

Behavior

𝑦 = 𝑠𝑖𝑛 (1

𝑥)

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 -4.99

f(x) 3.396 3.399 3.4 3.398 3.395

1. Use the graph of f(x) to the right to find

lim𝑥→−3

2𝑥2 + 7𝑥+3

𝑥+3

2. Use the table below to find lim𝑥→2

𝑔(𝑥)

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. lim𝑥→𝑎−

𝐻(𝑥) = lim𝑥→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 Analytically

Properties of Limits

Some Basic Limits

Let 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 Limits

1. Direct substitution

2. Factor, cancellation technique

3. The conjugate method, rationalize the numerator

4. Use special trig limits of lim𝑥→0

sin 𝑥

𝑥= 1 or lim

𝑥→0

1−cos 𝑥

𝑥= 0

Direct Substitution

1. lim𝑥→2

(3𝑥 − 5)

2. lim𝑥→4

√𝑥 + 43

3. lim𝑥→1

sin𝜋𝑥

2

4. lim𝑥→7

𝑥

5. If lim𝑥→𝑐

𝑓(𝑥) = 7 then lim𝑥→𝑐

5𝑓(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+1

𝑥+1

10. lim𝑥→2

𝑥3+1

𝑥+1

11. lim𝑥→−1

𝑥3+1

𝑥+1

Limits of Functions Involving a Radical

12. lim𝑥→3

√𝑥+1−2

𝑥−3

Dividing out Technique

13. lim∆𝑥→

2(𝑥+ ∆𝑥)−2𝑥

∆𝑥

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

Find limℎ→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

𝑥 cos 𝑥

19. lim𝑥→0

tan 𝑥

𝑥

20. lim𝑥→0

sin 3𝑥

𝑥

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 lim𝑥→𝑐

ℎ(𝑥) = 𝐿 = lim𝑥→𝑐

𝑔(𝑥) then lim𝑥→𝑐

𝑓(𝑥) exists and is equal to L.

4 – |𝑥| < f(x) < 4 + |𝑥|

Special Trigonometric Limits:

lim𝑥→0

sin 𝑥

𝑥= 1 lim

𝑥→0

1−cos 𝑥

𝑥= 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. lim𝑥→𝑐

𝑓(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 ° 𝑓

2. Sum or difference: f± g

3. Product: f • g

4. Quotient: 𝑓

𝑔 , if g(c) ≠ 0

5. 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

lim𝑥→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 lim𝑥→𝑎+

𝑓(𝑥) = 𝑓(𝑎) and lim𝑥→𝑏−

𝑓(𝑥) = 𝑓(𝑏)

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

3 ; 𝑥 = −2

𝑥3 − 6𝑥 + 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) = 𝑥2−4

𝑥+2 when x ≠ -2, then

f(-2) = ?

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

a. y = 𝑥2

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? 𝑔(𝑥) = {𝑐𝑥 + 1 ; 𝑖𝑓 𝑥 ≤ 3

𝑐𝑥2 − 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 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 – 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 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

lim𝑥→𝑐

𝑔(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

𝑐𝑔(𝑥)

𝑓(𝑥)= 0

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

1

𝑥2−9

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 − 1

𝑥)

Example 7: Evaluate lim𝑥→(

−1

2)

+

6𝑥2+𝑥−1

4𝑥2−4𝑥−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