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.