Continuity of Functions Practice Questions

A function f(x) is said to be continuous at a point x = a if the limit of the function as (x) approaches (a) exists and is equal to the value of the function at that point.

Solved Examples

Example 1: For function f(x) defined as 

\bold{f(x) = \begin{cases} 4x, & \text{if } x < 2 \\ ~~~~8, & \text{if } x = 2 \\ 3x+2, & \text{if } x > 2\end{cases}}

Check the continuity at x =2.

Solution:

We will calculate the limits at x = 2

LHL = Iim_(x→2-) f(x) = Iim_(h→0) f(2-h)

⇒ LHL = Iim_(h→0)4×(2-h)

⇒ LHL = 4×(2-0)

⇒ LHL = 8

RHL = Iim_(x→0+) f(x) = Iim_{(h → 0)} f(0+h)

⇒ RHL = Iim_{(h → 0)}3×(2+h)+2

⇒ RHL = 3×(2+0)+2

⇒ RHL = 8

Example 2: Find the value of 'm' at which the function is continuous at x = 9

\bold{f(x) = \begin{cases}mx+5 & \text{if} ~x ≠ 9\\ 8x & \text{if} ~x = 9\end{cases}}

Solution:

lim_{x → 9-} f(x) = lim (h → 0) f(9 - h)
LHL = lim_{h → 0} [m × (9 - h) + 5]
LHL = 9m + 5

RHL = lim_{x → 9+} f(x)
⇒ RHL = lim_{h → 0} f(9 + h)
⇒ RHL = lim_{h → 0} [8 × (9 + h)]
⇒ RHL = 72

For a function to be continuous Left Hand Limit is equal to Right Hand Limit

Thus, 9m + 5 =72

⇒ m = 7.4444.....

For the function to be continuous the value of m should be 7.444.....

Example 3: Test the continuity at x = 1, for the following function.

\bold{f(x) = \begin{cases}\frac{x^2-3x+2}{x-1} & \text{if} ~x ≠ 1\\ 9x & \text{if} ~x = 1\end{cases}}

Solution:

To test the continuity of the function f(x) at x = 1, the following conditions must be checked:

1. f(1) is defined.
2. lim⁡_x→1 f(x) exists.
3. lim⁡_x→1 f(x) = f(1).

Step 1. Evaluate f(1):

Given that f(x) = 9x  when x = 1:
f(1) = 9⋅1 = 9
Thus, f(1) is defined and equals 9.

Step 2. Evaluate lim⁡_x→1 f(x)

For x ≠ 1, f(x) = \frac{x^2 - 3x + 2}{x - 1}
x = 1 gives undefined value (i.e., 0/0)

Factorize and simplify the function.
x² − 3x + 2 = (x − 1)(x − 2)

Thus, for x ≠ 1,
f(x)= \frac{(x−1)(x−2)}{x -1 }= x -2

Therefore, lim⁡_x→1 f(x) = lim⁡_x→1 (x − 2) = 1 − 2 = −1
The limit exists and equals −1.

Step 3. Compare lim⁡_x→1 f(x) and f(1)

• lim _x→1 ​f(x) = −1
• f(1) = 9
  Since −1 ≠ 9, the limit does not equal the function value at x = 1.

The function f(x) is discontinuous at x = 1 because lim_⁡x→1 f(x) ≠ f(1).

Example 4: \bold{f(x) = \begin{cases}\frac{sin(x-2)}{x^2-4} & \text{if} ~x ≠ 2\\ 0.25 & \text{if} ~x = 2\end{cases}}

Test the continuity at x = 2 of the above-mentioned function.

Solution:

If we put x = 2 in the function sin(x - 2) / (x² - 4) then the function is undefined. Therefore we need to factorize and simplify the function.

\lim_{x\to 2}f(x) = \lim_{x\to 2} \frac{sin(x - 2)}{x^2 - 4} \\ \Rightarrow \lim_{x\to 2} f(x) = \lim_{x\to 2} \frac{sin(x - 2) }{(x - 2)\times (x + 2)} \\ \Rightarrow \lim_{x\to 2} f(x) = \lim(x\to 2) \frac{1}{x + 2}\times \lim(x\to 2) \frac{sin(x - 2)}{x - 2} 

⇒ lim_x→2 ​f(x) = 1/4 = 0.25
⇒ lim_x→2 ​f(x) = f(2) = 0.25

Thus, the function is continuous at x = 2.

Example 5: Find the relation between m and n if the function is continuous at point x = -5

\bold{f(x) = \begin{cases}mx^2 & \text{if} ~x < -5\\ nx+5 & \text{if} ~x \geqslant -5\end{cases}}

Solution:

We will compute the Left Hand Limit and Right Hand Limit separately

LHL = lim_{x→−5 −} mx²
⇒ LHL = lim_h→0 m×(−5−h)²
⇒ LHL = 25m

RHL = lim _{x→−5 +} nx+5 
⇒ RHL = lim_h→0  n×(-5+h)+5 
⇒ RHL = -5n + 5

For a function to be continuous LHL = RHL

25m = -5n + 5 
⇒ 25m + 5n = 5
⇒ 5m + n = 1

Practice Question

Question 1: Check whether f(x) = 3x + 2 is continuous at x = 1.

Question 2: Determine whether the function is continuous at x = 3: f(x)=\frac{x^2-9}{x-3}, \quad x\ne 3

Question 3: Find the value of k for which the function is continuous at x = 1:

f(x)= \begin{cases} x^2+k, & x<1\\ 4, & x=1\\ 2x+k, & x>1 \end{cases}

Question 4: Check whether the function is continuous at x = 1:

f(x)=\frac{x^2+x-2}{x-1}, \quad x\ne 1

Question 5: Determine the points of discontinuity of the function: f(x)=\frac{x+2}{x^2-4}

Answer Key

1. Continuous at x = 1
2. The function is undefined at x = 3; not continuous there
3. No such value of k exists
4. The function is undefined at x = 1; not continuous there
5. x = -2and x = 2