**Theorem, from Definition of Derivative**

If f(x) is differentiable at x = a then f(x) is continuous at x =a.

**Proof : **Since f(x) is differentiable at x = a we know,

f'(a) = lim _{x→a} (f(x) - f(a))/(x - a)

exists. We will require this in some.

If we next suppose that x ≠ a we can write the as given below,

f(x) - f(a) = ((f(x) - f(a))/( x -a)) (x -a)

Afterward fundamental properties of limits tells us as we have,

lim _{x→a} (f(x) - f(a)) = lim _{x→a }[((f(x) - f(a))/(x - a)) (x -a)]

= lim _{x→a} (f(x) - f(a))/(x - a) lim _{x→a }(x -a)

The primary limit on the right is only f′(a) as we considered above and the second limit is obviously zero and therefore,

lim _{x→a} (f(x) - f(a)) = f'(a).0 = 0

So we've managed to prove as,

lim _{x→a} (f(x) - f(a)) = 0

Although just how does this help us to x= a, prove that f(x) is continuous at x = a?

Let's establish with the subsequent.

lim _{x→a} (f(x)) = lim _{x→a} [f(x) + f(a) - f(a)]

Remember that we have just added in zero upon the right side. Some rewriting and the utilize of limit properties provides,

lim_{x→a} (f(x)) = lim_{x→a} [f(a) + f(x) - f(a)]

= lim_{x→a} f(a) + lim_{x→a} [f(x) - f(a)]

Here, we only proved above that lim_{x→a} [f(x) - f(a)] = 0 and since f(a) is a constant we also know that lim_{x→a} f(a) = f(a), then it should be,

lim_{x→a} f(x) = lim_{x→a} f(a) = 0 = f(a)

Or conversely, lim_{x→a} f(x) = f(a) although it is exactly what this means for f(x) is continuous at x = a and therefore we are done.