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Continuity of a Function

Continuity of a function says that a function 'f' is continuous at x = c means that there is no interruption in graph of f at c. That means the graph is unbroken at x= c and there are no holes, jumps or gaps.A function is continuous at x = c, if it satisfies the following three conditions.

1) f(c) is defined

2) $\lim_{x->c}f(x)$ exists.

3) $\lim_{x->c}f(x)$ = f(c)

A function is continuous at open interval (a,b) when the function is continuous at each point in the interval. A function that is continuous on the entire real number line (-$\infty $,$\infty $) is everywhere continuous.

## Examples on continuity of a function

1) Discuss the continuity of a given function f(x) = sin(x)**Solution :**the domain of f(x) is all real numbers. As we know that $\lim_{x->c}sin(x)$ = sin(c) for any real number 'c', so we conclude that the function is continuous on its entire domain (-$\infty$, $\infty$)

2) Discuss the continuity of a function f(x) = 3 - $\sqrt{9-x^{2}}$ on the closed interval[-3,3].

**Solution :**In the given function, there is a square root and we can not have negative radical under the square root. Find the value of 'x', that gives the radical greater or equal to zero.

$\sqrt{9-x^{2}} \geq 0 $

9 - $x^{2}$ = 0 ( Solve this like a simple equation)

$- x^{2}$ = -9

$ x^{2}$ = 9

x = $\pm $ 3

So x = 3 or x = -3 are the critical values.

Now set the intervals as (-$\infty$, -3], [-3,3],[3,$\infty$)

Let us consider x = -4 from the interval (-$\infty$, -3],

$\sqrt{9-(-4)^{2}} \geq 0 $

$\sqrt{9-16} \geq 0 $

-7 $\geq$ 0 is false.

Let us consider x = 0 from the interval [-3,3],

$\sqrt{9-(0)^{2}} \geq 0 $

$\sqrt{9-0} \geq 0 $

9 $\geq$ 0 is true.

Let us consider x = 4 from the interval [3,$\infty$),

$\sqrt{9-(4)^{2}} \geq 0 $

$\sqrt{9-16} \geq 0 $

-7 $\geq$ 0 is false.

So, f(x) = 3 - $\sqrt{9-x^{2}}$ is continuous at the closed interval [-3,3].

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