Prove that is everywhere continuous.

Basic Idea:

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, using our fundamental idea of limit from class 11 we can summarise it as, A function is continuous at x = c if :

Here we have,

…….equation 1

To prove it everywhere continuous we need to show that at every point in domain of f(x) [ domain is nothing but a set of real numbers for which function is defined ]

Clearly from definition of f(x) { see from equation 1}, f(x) is defined for all real numbers.

∴ we need to check continuity for all real numbers.

Let c is any random number such that c < 0 [thus c being a random number, it can include all negative numbers ]

f(c) =

[ using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x < 0

Now, let m be any random number from domain of f such that m > 0

thus m being a random number, it can include all positive numbers]

f(m) = m+1 [using eqn 1]

Clearly,

∴ We can say that f(x) is continuous for all x > 0

As zero is a point at which function is changing its nature so we need to check LHL, RHL separately

f(0) = 0+1 = 1 [using eqn 1]

LHL = [∵ sin –θ = sin θ and ]

RHL =

Thus LHL = RHL = f(0).

∴ f(x) is continuous at x = 0

Hence, we proved that f is continuous for x < 0 ; x > 0 and x = 0

Thus f(x) is continuous everywhere.

Hence, proved.

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