Limits exercises with answers

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. 2. 3.

f (−1)

5.

f (1)

9. lim f (x)

lim f (x)

6. lim− f (x)

10. lim− f (x)

x→−1−

lim f (x)

x→1

x→−1+

7. lim+ f (x)

4. lim f (x)

8. lim f (x)

x→ −1

x→1

x→1

x→3

x→5

Learning Activity 5

MthSc 105

Name: __________ Id:

__________

1. Assume lim f (x) = 8 and lim g(x) = 2. Compute the following limit by applying the Limit x→1

x→1

Laws.

 f (x)  lim   x→1 g(x)  

Find the limits. Simplify as much as possible. Use proper notation. 2. lim (10 − 3x) x→12

3. lim (2z − 8)1/ 3 z→0

Learning Activity 5

4. lim

h

h→0

3h + 1 − 1

5. lim

t 2 + 2t − 3 2t 2 + 3t − 5

t→1

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.

d.

x→1−

b. lim+ f (x)

c. lim f (x)

lim f (x)

e.

f.

lim f (x)

x→2−

x→1

x→1

lim f (x)

x→2+

lim f (x) x→2

Learning Activity Evaluate the limits or state that they do not exist (DNE). 2. lim

x 3 − 5x 2 x2

3.

lim+

2 (x − 3)3

lim−

2 (x − 3)3

x→0

x→3

4.

x→3

x2 . x 3 − 2x 2 + x For each vertical asymptote x = a, evaluate lim+ f (x) , lim− f (x) , and lim f (x) .

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

x→ a

x→ a

Learning Activity 7 1. Evaluate the limits.

 1 10  a. lim  5 + + 2  x→ ∞  x x 

b.

(

lim −3x16 + 2

x→ −∞

)

3x 4 + 3x 3 − 36x 2 x→∞ x 4 − 25x 2 + 144

c. lim

d. lim

x→∞

e.

lim

4x 3 2x 3 + 9x 6 + 15x 4

x→−∞

4x 3 2x 3 + 9x 6 + 15x 4

MthSc 105

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 lim− f ( x ) and lim+ f ( x ) . x→ a

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

x→ a

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 is continuous at a = 4. x − 9x + 20 2

3. Classify the discontinuities in the function h(x) = at the points x = 0 and x = 1.

x 3 − 4x 2 + 4x x(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. 2. 3.

f (−1) = 1

5.

f (1) = 4

9. lim f (x) = 4

lim f (x) = 2

6. lim− f (x) = 3

10. lim− f (x) = 5

x→−1−

lim f (x) = 2

x→−1+

4. lim f (x) = 2 x→−1

x→1

x→3

x→5

7. lim+ f (x) = 5 x→1

8. lim f (x) DNE because lim− f (x) ≠ lim+ f (x) x→1

x→1

x→1

Learning Activity 5

SOLUTION

MthSc 105

1. Assume lim f (x) = 8 and lim g(x) = 2. Compute the following limit by applying the Limit x→1

x→1

Laws.

f (x) 8  f (x)  lim x→1 lim  = = =4  x→1 g(x) g(x) 2   lim x→1 Find the limits. Simplify as much as possible. Use proper notation. 2. lim (10 − 3x) x→12

= 10 − 3(12) = −26

3. lim (2z − 8)1/ 3 z→0

= 3 2(0) − 8 = 3 −8 = −2

h

4. lim

3h + 1 − 1

h→0

( 3h + 1 + 1) 3h + 1 − 1 3h + 1 + 1 ( 3h + 1) − 1 h ( 3h + 1 + 1) h ( 3h + 1 + 1) = lim = lim = lim

h

h→0

h→0

= lim h→0

⋅

3h + 1 − 1

3h + 1 + 1 3

3(0) + 1 + 1 2 = 3 3

=

t 2 + 2t − 3 t→1 2t 2 + 3t − 5 (t + 3) (t − 1) = lim t→1 (2t + 5) (t − 1)

5. lim

t+3 t→1 2t + 5 1+ 3 = 2(1) + 5 = lim

=

4 7

3h + 1 + 1

= lim h→0

h→0

3h

h

2

2

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.

d.

x→1−

lim f (x) = ∞

b. lim+ f (x) = ∞

lim f (x) = ∞

e.

x→2−

c. lim f (x) = ∞ x→1

x→1

lim f (x) = −∞

x→2+

f.

lim f (x) DNE x→2

Learning Activity 6

SOLUTION

Evaluate the limits or state that they do not exist (DNE). 2. lim x→0

x 3 − 5x 2 x2

= lim

x 2 (x − 5)

x→0

3.

lim+

2 =∞ (x − 3)3

 approaches 2 2  positive and approaches 0 ≡ 0+ 

lim−

2 = −∞ (x − 3)3

 2 approaches 2  negative and approaches 0 ≡ 0− 

x→3

4.

= 0 − 5 = −5

x2

x→3

x2 . x 3 − 2x 2 + x For each vertical asymptote x = a, evaluate lim+ f (x) , lim− f (x) , and lim f ( x ) .

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

f (x) =

x→ a

x→ a

x2 x2 x 21 x = = = , where x ≠ 0 3 2 2 x − 2x + x x(x − 2x + 1) x (x − 1)2 (x − 1)2

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

x =∞ (x − 1)2

 1 approaches 1  positive and approaches 0 ≡ 0+ 

lim− f (x) = lim−

x =∞ (x − 1)2

 approaches 1 1  positive and approaches 0 ≡ 0+ 

x →1

x →1

x →1

x →1

lim f ( x ) = ∞ x →1

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

(lim f (x) = lim f (x) = ∞) x→1+

x→1−

SOLUTION

Learning Activity 7

1. Evaluate the limits.

 1 10  a. lim  5 + + 2  = 5 + 0 + 0 = 5 x→∞  x x  b.

(

)

lim −3x16 + 2 = −3⋅ ∞ + 2 = −∞ + 2 = −∞

x→−∞

3x 4 + 3x 3 − 36x 2 c. lim 4 x→∞ x − 25x 2 + 144 3x 4 + 3x 3 − 36 x 2 3 + 3 − 36 2 4 4 4 x x x x x = lim 4 = lim 2 x→ ∞ x x→∞ 25 144 1− + − 25x 4 + 144 4 x2 x4 x4 x x 3+ 0 − 0 = =3 1− 0 + 0 d. lim

x→∞

4x 3 2x 3 + 9x 6 + 15x 4 4x 3

= lim

x→ ∞

2x 3 x

=

e.

6 + 9x

3

4 2+ 9+0

lim

x→−∞

=

x3 x

6

+ 15x

= lim

4

x→∞

x

6

4 2 + 9 + 15

x2

4 4 = 2+3 5

4x 3 2x 3 + 9x 6 + 15x 4 4x 3

−4 − x3 = lim x→ −∞ x→−∞ 2x 3 9x 6 + 15x 4 −2 + 9 + 15 2 3 + 6 6 x −x x x −4 −4 = = = −4 −2 + 9 + 0 −2 + 3 = lim

MthSc 105

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. ½x –1 x2 + 0x – 1 x2 + 2x – 2x – 1 – 2x – 4 3

2x + 4

OA: y =

x2 − 1 3 = ½x–1+ 2x + 4 2x + 4 3 AND lim = 0 x→ ±∞ 2x + 4

Check: y =

1 x −1 2

b. Find the vertical asymptote(s) of f. For each vertical asymptote x = a, evaluate lim− f (x) and lim+ f (x) . x→ a

x→ a

VA: x = –2 x2 − 1 lim f (x) = lim− = −∞ x→−2− x→−2 2x + 4 x2 − 1 lim+ f (x) = lim+ =∞ x→−2 x→−2 2x + 4

3 0− 3 SCRATCH: + 0 SCRATCH:

+ . − + So, . + So,

c. Sketch f (x) and all of its asymptotes. x = –2

y y=

x2 − 1 2x + 4

y=½x–1 x

Learning Activity 8

SOLUTION

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:

lim f (x) ≠ f (1) x→1

(lim f (x) = 2, f (1) = 3) x→1

REMOVABLE x = 2:

lim f (x) DNE x→2

(lim f (x) = 1, lim f (x) = 2 x→2−

x→ 2+

JUMP x = 3:

f (3) is undefined REMOVABLE

⇒ lim− f (x) ≠ lim+ f (x) x→ 2

x→ 2

)

Learning Activity 8

SOLUTION

2. Determine if the function f (x) =

5x − 2 is continuous at a = 4. x − 9x + 20 2

(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 18 18 = = 4 − 9(4) + 20 16 − 36 + 20 0 2

So, f (4) is undefined.

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

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

x 3 − 4x 2 + 4x x(x − 1)

at the points x = 0 and x = 1.

(

) (

)

2

2 x x−2 x 3 − 4x 2 + 4x x x − 4x + 4 h(x) = = = x(x − 1) x(x − 1) x(x − 1)

Note the common factor of x in numerator and denominator. lim h(x) = lim x→ 0

x→ 0

x (x − 2)2 x (x − 1)

=

(0 − 2)2 = −4 0 −1

So, the discontinuity at x = 0 is REMOVABLE. Note that the denominator is zero when x = 1, but the numerator is not zero. => VA

lim+ h(x) = lim+

x→1

x→1

(x − 2)2 =∞ x −1

SCRATCH:

So, the discontinuity at x = 1 is INFINITE.

1 + → + + 0