Example 1 :
Find the points of discontinuity of the function f, where
Solution :
For the values of x greater than 3, we have to select the function 4x + 5.
lim x->3- f(x) = lim x->3- 4x + 5
= 4(3) + 5
= 12 + 5
= 17 -------(1)
For the values of x lesser than 3, we have to select the function 4x - 5.
lim x->3+ f(x) = lim x->3+ 4x - 5
= 4(3) - 5
= 12 - 5
= 7 -------(2)
lim x->3- f(x) ≠ lim x->3+ f(x)
So, the given piece-wise function is not continuous at
x = 3
Example 2 :
Find the points of discontinuity of the function f, where
Solution :
For the values of x greater than 2, we have to select the function x + 2.
lim x->2- f(x) = lim x->2- x + 2
= 2 + 2
= 4 -------(1)
For the values of x lesser than 2, we have to select the function x2.
lim x->2+ f(x) = lim x->2+ x2
= 22
= 4-------(2)
lim x->2- f(x) = lim x->2+ f(x)
The function is continuous at x = 2.
So, the given piece-wise function is continuous for all real values of x.
That is,
x ∈ R
Example 3 :
Find the points of discontinuity of the function f, where
Solution :
For the values of x greater than 2, we have to select the function x2 + 1
lim x->2- f(x) = lim x->2- x2 + 1
= 22 + 1
= 5 -------(1)
For the values of x lesser than 2, we have to select the function x3 - 3.
lim x->2+ f(x) = lim x->2+ x3 - 3
= 23 - 3
= 8 - 3
= 5-------(2)
lim x->2- f(x) = lim x->2+ f(x)
The function is continuous at x = 2.
So, the given piece-wise function is continuous for all real values of x.
That is,
x ∈ R
Example 4 :
Find the points of discontinuity of the function f, where
Solution :
Here we are going to check the continuity between 0 and π/2.
For the values of x lesser than or equal to π/4, we have to choose the function sin x.
lim x->π/4- f(x) = lim x->π/4- sin x
= sin (π/4)
= 1/√2
For the values of x greater than π/4, we have to choose the function cos x .
lim x->π/4+ f(x) = lim x->π/4+ cos x
= cos (π/4)
= 1/√2
The function is continuous for all x ∈ [0, π/2).
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