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, using our fundamental idea of the limit from class 11 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,

\(f(x) = \begin{cases}\frac{sin\,x}{x}&,\quad x <0\\x+1&,\quad x≥0\end{cases} \) …….equation 1

To prove it everywhere continuous we need to show that at every point in the domain of f(x) 

[domain is nothing but a set of real numbers for which function is defined ]

 \(\lim\limits_{x \to c}f(x)\) = f(c),

Where c is any random point from domain of f.

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) = \(\frac{sin\,c}{c}\) [ using eqn 1]

\(\lim\limits_{x \to c}f(x)\) =\(\lim\limits_{x \to c}\frac{sin\,x}{x}\) = \(\frac{sin\,c}{c}\)

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

Now, 

Let m be any random number from the 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]

\(\lim\limits_{x \to m}f(x)\) = \(\lim\limits_{x \to m}x+1\) = m + 1

Clearly,

\(\lim\limits_{x \to c}f(x)\) = f(c) = m + 1

∴ 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 = \(\lim\limits_{h \to 0}f(0-h)\) 

= \(\lim\limits_{h \to 0}\frac{sin\,(-h)}{-h}\) 

= \(\lim\limits_{h \to 0}\frac{sin\,h}{h}\) = 1

[∵ sin – θ = – sin θ and \(\lim\limits_{h \to 0}\frac{sin\,h}{h}\) = 1 ]

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

= \(\lim\limits_{h \to 0}h+1\) = 1

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