CHECKING TYPES OF DISCONTINUITY

State whether or not each of the following functions is continuous.

• If not, state where the discontinuity occurs and whether or not it is removable.
• Is the discontinuity is an asymptote, a hole or jump ?
• If an asymptote, what is its equation ?

Problem 1 :

Solution :

For any real values of x, the function will not become undefined. So, it is continuous for all real values.

Problem 2 :

Solution :

Equating the denominators to zero, we get

The function is discontinuous at x = -1/2 and 1.

Type of discontinuity = Non removable

Vertical asymptotes are at x = -1/2 and x = 1.

Problem 3 :

Solution :

Equating the denominators to 0, we get

The function is discontinuous at x = 3 and -2.

Type of discontinuity = Non removable

Vertical asymptotes are at x = 3 and x = -2.

Problem 4 :

Solution :

It is not continuous at x = 4 and x = -4. Removing the common factor, we get f(x) = 1/(x + 4)

Equating the denominator to 0, we get

x + 4 = 0

x = -4

So, the vertical asymptote is at x = -4

Hole is at x = 4. When x = 4, y = 1/5

Problem 5 :

Solution :

The function is not continuous at x = 3, there is removable discontinuity at (3, 6).

Problem 6 :

Solution :

Lim _x->2^-  f(x) = Lim _x->2- 2x - 3

Applying the limit, we get

Lim _x->2^-  f(x) = 2(2) - 3

= 4 - 3

= 1

Lim _x->2⁺  f(x) = Lim _x->2+ x²

Applying the limit, we get

Lim _x->2⁺  f(x) = 2²

= 4

Since Lim _x->2^-  f(x) = Lim _x->2⁺  f(x), then lim _{x ->2} f(x) does not exists.

Type of discontinuity = Jump discontinuity

Problem 7 :

Solution :

Lim _x->-1^-  f(x) = Lim _x->-1^- x³

Applying the limit, we get

Lim _x->-1^-  f(x) = (-1)³

= -1 -----(1)

Lim _x->-1⁺  f(x) = Lim _x->-1⁺ x

Applying the limit, we get

Lim _x->-1⁺  f(x) = -1-----(2)

(1) = (2)

At x = -1, the function is continuous.

Lim _x->1^-  f(x) = Lim _x->1^- x

Applying the limit, we get

Lim _x->1^-  f(x) = 1 -----(1)

Lim _x->1⁺  f(x) = Lim _{x->1^{+ (}}1 -  x)

Applying the limit, we get

Lim _x->1⁺  f(x) = 1 - 1

= 0 ----(2)

(1) ≠ (2)

At x = 1, the function is not continuous, then the limit does not exists.

There is jump discontinuity at x = 1.

Problem 8 :

Solution :

|x| - 3

When x is positive

x - 3

When x is negative

-x - 3

f(x) is not continuous at x = 3 and x = -3, there is vertical asymptote at x = 3 and x = -3

Problem 9 :

Find the value of a if the function is continuous.

Solution :

Lim _x->1^-  f(x) = Lim _x->1^- 7x - 2

Applying the limit, we get

Lim _x->1^-  f(x) = 7(1) - 2

= 5 ----(1)

Lim _x->1⁺  f(x) = Lim _x->1⁺ ax²

Applying the limit, we get

Lim _x->1⁺  f(x) = a(1)²

= a----(2)

(1) = (2)

a = 5

Problem 10 :

Find the value of a if the function is continuous.

Solution :

Lim _x->2^-  f(x) = Lim _x->2^- ax²

Applying the limit, we get

Lim _x->2^-  f(x) = a(2)²

= 4a ----(1)

Lim _x->2⁺  f(x) = Lim _x->2⁺ 2x + a

Applying the limit, we get

Lim _x->2⁺  f(x) = 2(2) + a

= 4 + a----(2)

(1) = (2)

4a = 4 + a

4a - a = 4

3a = 4

a = 4/3

Problem 11 :

Solution :

lim _x->1^-  f(x) = Lim _x->1^- x + 1

Applying the limit, we get

Lim _x->1^-  f(x) = 1 + 1

= 2

Lim _x->1⁺  f(x) = Lim _x->1⁺ ax + b

Applying the limit, we get

Lim _x->1⁺  f(x) = a(1) + b

= a + b

Equating left hand limit and right hand limit, we get

a + b = 2 ----(1)

lim _x->2^-  f(x) = Lim _x->2^- ax + b

Applying the limit, we get

Lim _x->2^-  f(x) = a(2) + b

= 2a + b

Lim _x->2⁺  f(x) = Lim _x->2⁺ 3x

Applying the limit, we get

Lim _x->2⁺  f(x) = 3(2)

= 6

Equating the left hand limit and right hand limit, we get

2a + b = 6 ----(2)

(1) - (2)

a + b - (2a + b) = 2 - 6

-a = -4

a = 4

Applying the value of a in (1), we get

2(4) + b = 6

b = 6 - 8

b = -2