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Continuity of Special Types of Functions

Continuity of functions in which greatest integer function is involved f(x) = [x] is discontinuous when x is an integer.
 
Similarly f(x) = [g(x)] is discontinuous when g(x) is an integer, but this is true only when g(x) is monotonic (g(x) is strictly increasing or strictly decreasing).
 
For example, f(x) = 88070.png is discontinuous when 88064.png is an integer, as 88058.png is strictly increasing (monotonic function).
 
f(x) = [x2], x ≥ 0 is discontinuous when x2 is an integer, as x2 is strictly increasing for the x ≥ 0.
 
Now consider f(x) = [sin x], x [0, 2π] function. g(x) = sin x is not monotonic in [0, 2π]. For this type of function points of discontinuity can be determined easily by graphical methods. We can note that at x = 3π/2, sin x takes integral value –1, but at x = 3π/2, f(x) = [sin x] is continuous.
 
Continuity of functions which signum function is involved We know that f(x) = sgn(x) is discontinuous at x = 0.
 
In general, f(x) = sgn(g(x)) is discontinuous at x = a if g(a) = 0.
 
Continuity of functions involving 88052.png We know that 88046.png
 
Illustration
Discuss continuity of f(x) = 88040.png.
Solution
f(x) = 88034.png = 88028.png
88020.png = 88014.png
 
Thus f(x) is discontinuous at x = ±1.
 
 
Continuity of functions in which f(x) is defined differently for rational and irrational values of x
 
Illustration
Discuss the continuity and discontinuity of the following function:
f(x) = 88008.png
Solution
For any x = a,
 
LHL = 88002.png = 0 or 1
 
(as 87996.png can be rational or irrational)
 
Similarly, RHL = 87990.png = 0 or 1
 
Hence f(x) oscillates between 0 and 1 as for all values of a.
 
Therefore, LHL and RHL do not exist.
 
Hence f(x) is discontinuous at a point x = a for all values of a.
 
 
Illustration
Find the value of x where f(x) = 87984.png is continuous.
 
Solution
f(x) is continuous at some x = a, where x = 1 – x or x = 1/2.
 
Hence f(x) is continuous at x = 1/2.
 
Explanation:
 
We have f(1/2) = 1/2.
 
If x → 1/2+ then x may be rational or irrational
⇒ f(1/2+) = 1/2 or 1 – 1/2 = 1/2.
 
If x → 1/2 then x may be rational or irrational
⇒ f(1/2) = 1/2 or 1 – 1/2 = 1/2.
 
Hence f(x) is continuous at x = 1/2. For some other point say x = 1.
 
f(1) = 1
 
If x → 1+ then x may be rational or irrational
⇒ f(1+) = 1 or 1 – 1 = 0.
 
Hence f(1+) oscillates between 1 and 0, which causes discontinuity at x = 1.
 
Similarly, f(x) oscillates between 0 and 1 for all x ∈ R – {1/2}.
 




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