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Prove that \(f(x) = \begin{cases} \frac{sin\,x}{x}&, \quad \text{if } x <{0}\\ x+1& ,\quad \text{if } x≥{ 0} \end{cases} \) is everywhere continuous.

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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,

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

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 ]

∴ 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]

∴ 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]

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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