Book: RD Sharma - Mathematics (Volume 1)
Chapter: 9. Continuity
Subject: Maths - Class 12th
Q. No. 14 of Exercise 9.1
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Examine the continuity of the function 
at x = 0. Also sketch the graph of this function.
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
If:–
equation 1
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 i.e. 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.
Let’s Solve now:
Given function is
…… Equation 2
We need to check whether f(x) is continuous at x=0 or not
For this we need to check LHL, RHL and value of function at x=0
Clearly,
f(0) = 3*0 – 2 = –2 [from equation 2]
LHL = 
RHL = 
As, LHL ≠ RHL
∴ f(x) is discontinuous at x = 0
This can also be proved by plotting f(x) on cartesian plane.
For x >0 ,we need to plot
y = x + 1
put y=0, we get x=–1 and for second point we put x=0 and thus get y=1
two points are enough to plot the straight line.
Two coordinates are (–1,0) and (0,1)
For x≤0, we need to plot
y = 3x – 2
put x = 0 then y = –2
on putting y=0 we get x = 2/3
two coordinates are (0,–2) and (
)
Graph:
It can be seen from graph that there is breakage in curve at (0,0)
Thus, it is discontinuous at x = 0
