Show that is discontinuous at x = a.
Ideas required to solve the problem:
1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.
Mathematically we define the same thing as given below:
A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked
where h is a very small positive no (can assume h = 0.00000000001 like this )
It means :–
Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c f(c).
Thus, it is the necessary condition for a function to be continuous
So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.
2. Idea of modulus function |x| : You can think this function as a machine in which you can give it any real no. as an input it returns it absolute value i.e if positive is entered it returns the same no and if negative is entered it returns the corresponding positive no.
Eg:– |2| = 2 ; |–2| = –(–2) = 2
Similarly, we can define it for variable x, if x ≥ 0 |x| = x
If x < 0 |x| = (–x)
Now we are ready to solve the question –
We need to check the continuity at x = a
So, we need to see whether at x=a,
LHL = RHL = f(a)
For this question c = a
Using the concept of modulus function we can redefine the above equation as given below :
f(x) = ……Equation 2
f(a) = 1
LHL = = using eqn 2
RHL = using eqn 2
Clearly, LHL ≠ RHL
∴ We can easily say that f(x) is discontinuous at origin.