Given that function f(x) is derivable at x = c

i.e. f'(x) \(=\underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}\) exists ........(1)

Now, we have to prove that f(x) is continuous function at x = c

i.e. we have to prove that \(\underset{x\rightarrow c}{lim}\) f(x) = f(c)

Now,  \(\underset{x\rightarrow c}{lim}\) f(x) - f(c) =  \(\underset{x\rightarrow c}{lim}\) (f(x) - f(c)) \((\frac{x-c}{x-c})\)

(By multiply numerator and denominator by x - c)

\(=(\underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c})\) \(\underset{x\rightarrow c}{lim}\) (x - c)  [∵ \(\underset{x\rightarrow c}{lim}\) f₁(x) f₂(x) = {\(\underset{x\rightarrow c}{lim}\) f₁(x)} {\(\underset{x\rightarrow c}{lim}\) f₂(x)}]

= f'(c) \(\underset{x\rightarrow c}{lim}\) (x - c)  (From equation (1))

= f'(c) x (c - c) = 0  (By taking limit)

Hence, (\(\underset{x\rightarrow c}{lim}\) f(x) - f(c)) = 0

⇒ \(\underset{x\rightarrow c}{lim}\) f(x) - \(\underset{x\rightarrow c}{lim}\) f(c) = 0

⇒  \(\underset{x\rightarrow c}{lim}\) f(x) - f(c) = 0

⇒  \(\underset{x\rightarrow c}{lim}\) f(x) = f(c)

Which implies that function f(x) is continuous at x = c.

Thus, we can say that if a function is derivable at a point then it is continuous at that point.

(∵ c is an orbitrary point)