Continuity of a function

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous.

Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.
let’s consider some examples that fail to meet our intuitive notion of what it means to be continuous at a point.

\( f(x)=\frac{x^2-1}{x-1}\)	\( \small {f(x)= \begin{cases} x+1 & \text{for } x ≤ 2 \\ x+2 & \text{for } x > 2 \end{cases}} \)	\( \small { f(x)= \begin{cases} x+1 & \text{for } x \ne 2 \\ 4 & \text{for } x = 2 \end{cases}} \)

1. In Figure (1). We see that the graph of f(x) has a hole at a. In fact, f(a) is undefined. So f(x) is NOT contnuous
2. In Figure (2), f(a) is defined, but the function has a jump at a. So f(x) is NOT contnuous.
3. In Figure (3), f(a) is defined, but the function has a gap at a. So f(x) is NOT contnuous.

Definition

A function f(x) is defined to be continuous at a if the following three conditions holds

1. f(a) is defined
2. \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
3. \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
4. LHL = RHL = f(a)

Examples

1. A function \( f(x) = x^2 + 1 \) is continuous at 2, since
   \( \displaystyle \lim_{x \rightarrow 2} f(x) =5 =f(2)\)

2. A function \( f(x) = \sqrt{4-x^2} \) is NOT continuous at 3
   since f(3) is not defined.

3. A function \( f(x)=\frac{1}{x-2}\)
   is not continuous at 2 because f(2) is not defined.

NOTES

1. The constant function, f (x) = c, \( \forall x \in R\) is continuous on R.
2. The identity function, f (x) = x, \( \forall x \in R\) is continuous on R.
3. The function f (x) = x n, \( x \in N\) is continuous on R.
4. The function f (x) = sinx is continuous.

Example 1

Test the continuity of a function \( f(x)= \begin{cases} -x^2 &\text { for } x \ne 2 \\ 0 &\text {for } x=2 \end{cases} \) at x=2.

Solution

The solution is as follows
At \( x=2^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2 =-4\) LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2 =-4\) RHL
At x=2, the value of the function is
f(2) = 0 Functional Value
Since, limiting value and functional value are not equal, f(x) is not continuous at x=2.

Example 2

Test the continuity of a function \( f(x)= \begin{cases} -x^2 +2 &\text { for } x \ne 2 \\ 1 &\text { for } x=2 \end{cases} \) at x=2.

Solution

The solution is as follows
At \( x=2^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2+2 =-2\) LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2+2 =-2\) RHL
At x=2, the value of the function is
f(2) = 1 Functional Value
Since, limiting value and functional value are not equal, f(x) is not continuous at x=2.

Example 3

Test the continuity of a function \(f(x)=\frac{|x|}{x}\) for \(x\ne 0\) at x=0.

Solution

The solution is
At x = 0, the value of f(0) is not defined.
Also, the left hand limit is -1, whereas the right hand limit is 1
So, the function is not continuous at x=0.

Example 4

Is\( f(x)=\frac{x+4}{x-3}\) continuous at x =1? At x =3?

Solution

The solution is as follows
At x=1, the limit value is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1}\frac{x+4}{x-3}=\lim_{x \to 1} \frac{1+4}{1-3} =\frac{5}{-2}\) This is Limit Value
At x=1, the value of the function is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1} \frac{1+4}{1-3} =\frac{5}{-2}\) This is Functional Value
Since, limiting value and functional value are equal, f(x) is continuous at x=1.
Next,
At x=3, the limit value is
\( \displaystyle \lim_{x \to 3} f(x)= \lim_{x \to 3} \frac{x+4}{x-3} =\frac{7}{0}\)=undefined
Since, the limiting value is undefined, f(x) is NOT continuous at x=3.