**Continuity at A Point**

We say a function f(x) is continuous at a point x = a it means at a point (a, f (a)). The graph of the function has no holes or graphs. That is, its graph is unbroken at point (a, f (a)).

The continuity at x = a₂ can be destroyed if \(\underset{x\to {{a}_{2}}}{\mathop{\lim }}\,f\left( x \right)\) does not exist.

**Continuity at A Point: **A function f(x) is said to be continuous at a point x = a of its domain if \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\).

Thus, f(x) is continuous at x = a \(\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\).

\(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\).

If f(x) is not continuous at a point x = a, then it is said to be discontinuous at x = a.

If \(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\ne f\left( a \right)\), then the discontinuity is known as the removable discontinuity because f(x) can be made continuous by re-defining it at point x = a in such a way that \(f\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\).

If \(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\), then f(x) is said to have a discontinuity of first kind.

A function f(x) is said to have a discontinuity of the second kind at x = a if \(\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\) or \(\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\) or both do not exist.

**Continuity on an Open Interval: **A function f(x) is said to be continuous on an open interval (a, b) if it is continuous at every point on the interval (a, b).

**Continuity on an Closed Interval: **A function f(x) is said to be continuous on a closed interval {a, b} if\(\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)\).

Continuous on the open interval (a, b) if \(\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( b \right)\).

In order words, f(x) is continuous on [a, b] if it is continuous on (a, b) and it is continuous at a form the right and at b form the left.

**Example: **Show that \(\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,(\frac{2|x|}{x}+x+1)=3\).

**Solution: **When x → 0⁺

|x| = x, x > 0

\(\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,(\frac{2x}{x}+x+1)\).

\(\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,(\frac{2}{1}+x+1)\).

(2 + 0 + 1) = 3.