Learning Activity 4 MthSc 105 Name: _________________________

Id: ___________________________ Use the graph of f in the figure to find the following values, if they exist. If a limit does not exist, answer “DNE” and explain why.

1. f (−1) 5. f (1) 9.

limx→3

f (x)

2.

lim

x→−1−f (x) 6.

limx→1−

f (x) 10. limx→5−

f (x)

3.

lim

x→−1+f (x) 7.

limx→1+

f (x)

4.

limx→−1

f (x) 8. limx→1

f (x)

Learning Activity 5 MthSc 105 Name: __________

Id: __________ 1. Assume

limx→1

f (x) = 8 and limx→1

g(x) = 2. Compute the following limit by applying the Limit

Laws.

limx→1

f (x)

g(x)

Find the limits. Simplify as much as possible. Use proper notation.

2. limx→12

(10 − 3x)

3. limz→0

(2z − 8)1/3

Learning Activity 5

4. limh→0

h

3h + 1 − 1

5. limt→1

t2 + 2t − 3

2t2 + 3t − 5

Learning Activity 6 MthSc 105 ________________________________________ 1. The graph of f in the figure has vertical asymptotes at x = 1 and x = 2. Find the

following limits, if possible.

a.

limx→1−

f (x) b. limx→1+

f (x) c. limx→1

f (x)

d.

limx→2−

f (x) e. limx→2+

f (x) f. limx→2

f (x)

Learning Activity

Evaluate the limits or state that they do not exist (DNE).

2. limx→0

x3 − 5x2

x2

3. limx→3+

2

(x − 3)3

4. limx→3−

2

(x − 3)3

5. Find all vertical asymptotes of the function f (x) =

x2

x3 − 2x2 + x.

For each vertical asymptote x = a, evaluate limx→a+

f (x) , limx→a−

f (x) , and limx→a

f (x) .

Learning Activity 7 MthSc 105

1. Evaluate the limits.

a. limx→∞

5 +1

x+

10

x2

b. limx→−∞

−3x16 + 2( )

c. limx→∞

3x4 + 3x3 − 36x2

x4 − 25x2 + 144

d. limx→∞

4x3

2x3 + 9x6 + 15x4

e. limx→−∞

4x3

2x3 + 9x6 + 15x4

Learning Activity

2. Sketch the rational function including all of its asymptotes.

y =

x2 − 1

2x + 4

Road map …

a. Use polynomial long division to find the oblique asymptote of f. b. Find the vertical asymptote(s) of f. For each vertical asymptote x = a, evaluate

limx→a−

f (x) and limx→a+

f (x) .

c. Sketch f (x) and all of its asymptotes.

Learning Activity 8 MthSc 105 ________________________________________ 1. Determine the values of x at which the function f graphed below has discontinuities.

For each value of x, state the condition(s) in the continuity checklist that are violated.

Learning Activity

2. Determine if the function f (x) =

5x − 2

x2 − 9x + 20 is continuous at a = 4.

3. Classify the discontinuities in the function h(x) =

x3 − 4x2 + 4x

x(x − 1)

at the points x = 0 and x = 1.

SOLUTION Learning Activity 4 MthSc 105

Use the graph of f in the figure to find the following values, if they exist. If a limit does not exist, answer “DNE” and explain why.

1. f (−1) = 1 5. f (1) = 4 9.

limx→3

f (x) = 4

2.

lim

x→−1−f (x) = 2 6.

limx→1−

f (x) = 3 10. limx→5−

f (x) = 5

3.

lim

x→−1+f (x) = 2 7.

limx→1+

f (x) = 5

4.

limx→−1

f (x) = 2 8. limx→1

f (x) DNE because limx→1−

f (x) ≠ limx→1+

f (x)

SOLUTION Learning Activity 5 MthSc 105

1. Assume limx→1

f (x) = 8 and limx→1

g(x) = 2. Compute the following limit by applying the Limit

Laws.

limx→1

f (x)

g(x)

=

limx→1

f (x)

limx→1

g(x)=

8

2= 4

Find the limits. Simplify as much as possible. Use proper notation.

2. limx→12

(10 − 3x)

= 10 − 3(12) = −26

3. limz→0

(2z − 8)1/3

= 2(0) − 83 = −83 = −2

4. limh→0

h

3h + 1 − 1

= limh→0

h

3h + 1 − 1⋅

3h + 1 + 1

3h + 1 + 1= lim

h→0

h 3h + 1 + 1( )3h + 1( )2

− 12

= limh→0

h 3h + 1 + 1( )3h + 1− 1

= limh→0

h 3h + 1 + 1( )3 h

= limh→0

3h + 1 + 1

3

=3(0) + 1 + 1

3=

2

3

5. limt→1

t2 + 2t − 3

2t2 + 3t − 5

= limt→1

(t + 3) (t − 1)

(2t + 5) (t − 1)

= limt→1

t + 3

2t + 5

=1+ 3

2(1) + 5

=4

7

SOLUTION Learning Activity 6 MthSc 105

1. The graph of f in the figure has vertical asymptotes at x = 1 and x = 2. Find the

following limits, if possible.

a.

limx→1−

f (x) = ∞ b. limx→1+

f (x) = ∞ c. limx→1

f (x) = ∞

d.

limx→2−

f (x) = ∞ e. limx→2+

f (x) = −∞ f. limx→2

f (x) DNE

Learning Activity 6 SOLUTION

Evaluate the limits or state that they do not exist (DNE).

2. limx→0

x3 − 5x2

x2

= lim

x→0

x2 (x − 5)

x2= 0 − 5 = −5

3. limx→3+

2

(x − 3)3= ∞

approaches 2

positive and approaches 0≡

2

0+

4. limx→3−

2

(x − 3)3= −∞

approaches 2

negative and approaches 0≡

2

0−

5. Find all vertical asymptotes of the function f (x) =

x2

x3 − 2x2 + x.

For each vertical asymptote x = a, evaluate limx→a+

f (x) , limx→a−

f (x) , and limx→a

f (x) .

f (x) =

x2

x3 − 2x2 + x=

x2

x(x2 − 2x + 1)=

x 2 1

x (x − 1)2=

x

(x − 1)2, where x ≠ 0

So, the denominator of f is zero AND the numerator is nonzero when x = 1.

limx→1+

f (x) = limx→1+

x

(x − 1)2= ∞

approaches 1

positive and approaches 0≡

1

0+

limx→1−

f (x) = limx→1−

x

(x − 1)2= ∞

approaches 1

positive and approaches 0≡

1

0+

limx→1

f (x) = ∞

limx→1+

f (x) = limx→1−

f (x) = ∞( ) Thus, x = 1 is a vertical asymptote of f.

SOLUTION Learning Activity 7 MthSc 105

1. Evaluate the limits.

a. limx→∞

5+1

x+

10

x2

= 5+ 0 + 0 = 5

b. limx→−∞

−3x16 + 2( )= −3⋅ ∞ + 2 = −∞ + 2 = −∞

c. limx→∞

3x4 + 3x3 − 36x2

x4 − 25x2 + 144

= limx→∞

3x4

x4 + 3x3

x4 − 36x2

x4

x4

x4 − 25x2

x4 + 144x4

= limx→∞

3+ 3x

− 36x2

1− 25x2 + 144

x4

=3+ 0 − 0

1− 0 + 0= 3

d. limx→∞

4x3

2x3 + 9x6 + 15x4

= limx→∞

4x3

x3

2x3

x3 + 9x6

x6 + 15x4

x6

= limx→∞

4

2 + 9 + 15x2

=4

2+ 9 + 0=

4

2+ 3=

4

5

e. limx→−∞

4x3

2x3 + 9x6 +15x4

= limx→−∞

4x3

−x3

2x3

−x3 + 9x6

x6 + 15x4

x6

= limx→−∞

−4

−2+ 9 + 15x2

=−4

−2 + 9 + 0=

−4

−2 + 3= −4

Learning Activity 7 SOLUTION

2. Sketch the rational function including all of its asymptotes.

y =

x2 − 1

2x + 4

Road map …

a. Use polynomial long division to find the oblique asymptote of f. ½ x – 1 2x + 4 x2 + 0x – 1 x2 + 2x – 2x – 1 – 2x – 4 3

OA: y =

1

2x − 1

b. Find the vertical asymptote(s) of f. For each vertical asymptote x = a, evaluate

limx→a−

f (x) and limx→a+

f (x) .

VA: x = –2

lim

x→−2−f (x) = lim

x→−2−

x2 − 1

2x + 4= −∞ SCRATCH:

3

0− So, +−

.

lim

x→−2+f (x) = lim

x→−2+

x2 − 1

2x + 4= ∞ SCRATCH:

3

0+ So, ++

.

c. Sketch f (x) and all of its asymptotes. y = ½ x – 1

x = –2

y = x2 − 1

2x + 4

x

y

Check: y = x2 − 1

2x + 4 = ½ x – 1 +

3

2x + 4

AND lim

x→±∞

3

2x + 4 = 0

SOLUTION Learning Activity 8 MthSc 105

1. Determine the values of x at which the function f graphed below has discontinuities.

For each value of x, state the condition(s) in the continuity checklist that are violated.

x = 1:

limx→1

f (x) ≠ f (1)

limx→1

f (x) = 2, f (1) = 3( )

REMOVABLE x = 2:

limx→2

f (x) DNE

limx→2−

f (x) = 1, limx→2+

f (x) = 2 ⇒ limx→2−

f (x) ≠ limx→2+

f (x)( )

JUMP x = 3: f (3) is undefined REMOVABLE

Learning Activity 8 SOLUTION

2. Determine if the function f (x) =

5x − 2

x2 − 9x + 20 is continuous at a = 4.

(We must determine if the function satisfies the three conditions for continuity at a = 4.)

1st condition: f (4) must be defined.

5(4) − 2

42 − 9(4) + 20=

18

16 − 36 + 20=

18

0 So, f (4) is undefined.

Thus, the function f is not continuous at a = 4.

3. Classify the discontinuities in the function h(x) =

x3 − 4x2 + 4x

x(x − 1)

at the points x = 0 and x = 1.

h(x) =

x3 − 4x2 + 4x

x(x − 1)=

x x2 − 4x + 4( )x(x − 1)

=x x − 2( )2

x(x − 1)

Note the common factor of x in numerator and denominator.

limx→0

h(x) = limx→0

x (x − 2)2

x (x − 1)=

(0 − 2)2

0 − 1= −4

So, the discontinuity at x = 0 is REMOVABLE.

Note that the denominator is zero when x = 1, but the numerator is not zero. => VA

limx→1+

h(x) = limx→1+

(x − 2)2

x − 1= ∞ SCRATCH:

1

0+ →++

So, the discontinuity at x = 1 is INFINITE.