A real function f is said to be continuous at x = c, where c is any point in the domain of f if :
where h is a very small ‘+ve’ no.
i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.
This is very precise,
We can summarise it as,
A function is continuous at x = c if :
Here we have,
\(f(x) = \begin{cases} -2&, \quad \text{if }\,x ≤{-1} \\2x&,\quad{if}\, -1<x<1\\2&,\quad{if}\,x≥1\end{cases} \) ….equation 1
Function is defined for all real numbers so we need to comment about its continuity for all numbers in its domain
(domain = set of numbers for which f is defined)
For x < –1, f(x) is having a constant value, so the curve is going to be straight line parallel to x–axis.
So, it is everywhere continuous for x < –1.
It can be verified using limits as discussed in previous problems
Similarly for –1 < x < 1,
Plot on X–Y plane is a straight line passing through origin.
So, it is everywhere continuous for –1 < x < 1.
And similarly for x > 1,
Plot is going to be again a straight line parallel to x–axis
∴ it is also everywhere continuous for x > 1
From graph it is clear that function is continuous everywhere but let’s verify it with limits also.
As x = –1 is a point at which function is changing its nature so we need to check the continuity here.
f(–1) = –2 [using eqn 1]
Thus,
LHL = RHL = f(–1)
∴ f(x) is continuous at x = –1
Also at x = 1
function is changing its nature so we need to check the continuity here too.
f(1) = 2 [using eqn 1]
Thus,
LHL = RHL = f(1)
∴ f(x) is continuous at x = 1
Thus,
f(x) is continuous everywhere and there is no point of discontinuity.
