A real function f is said to be continuous at x = c, where c is any point in the domain of f if :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

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  :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases} x^{10}&, \quad \text{if } x ≤{1} \\x^2&,\quad \text{if } x >{1} \end{cases} \) ……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)

Let c is any random number such that c < 1 

[thus c being random number, it is able to include all numbers less than 1]

f(c) = c¹⁰ [ using eqn 1]

\(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}x^{10}\)= c¹⁰

Clearly,

\(\lim\limits_{x \to c}f(x)\) = f(c)

∴ We can say that f(x) is continuous for all x < 1

As x = 1 is a point at which function is changing its nature so we need to check the continuity here.

f(1) = 1¹⁰ = 1 [using eqn 1]

LHL = \(\lim\limits_{h \to 0}f(1-h)\) 

= \(\lim\limits_{h \to 0}(1-h)^{10}\) = 1

RHL = \(\lim\limits_{h \to 0}f(1+h)\) 

= \(\lim\limits_{h \to 0}(1+h)^{2}\) = 1

Thus, 

LHL = RHL = f(1) 

∴ f(x) is continuous at x = 1 

Let m is any random number such that m > 1 

[thus m being random number, it is able to include all numbers greater than 1]

f(m) = m² [ using eqn 1]

And,

 \(\lim\limits_{x \to m}f(x)\) = \(\lim\limits_{x \to m}x^{2}\)=m²

Clearly,

\(\lim\limits_{x \to m}f(x)\) = f(m)

∴ We can say that f(x) is continuous for all m > 1 

Hence, 

f(x) is continuous for all real x 

There no point of discontinuity. It is everywhere continuous