# Continuity

The intuitive idea of a continuous function $ƒ$ in the interval $[a, b]$ gives the impression that the graph of the function $ƒ$ in this interval is a smooth curve without any break in it. Similarly, a discontinuous function gives the picture consisting of disconnected curves.

# Continuos and Discontinuous Functions

The function $ƒ(x)$ is said to be continuos at the point $x=x_o$, if and only if $\lim_{x \to x_o} ƒ(x)=ƒ(x_o)$ This definition of continuity of the function $ƒ(x)$ at $x=x_o$ implies that

1. $\lim_{x \to x_o} ƒ(x)$ exists i.e. $\lim_{x \to x_o^-} ƒ(x)$ and $\lim_{x \to x_o^+} ƒ(x)$ are finite and equal.
2. $ƒ(x_o)$ exists.
3. $\lim_{x \to x_o} ƒ(x)=ƒ(x_o)$

Hence, $ƒ(x)$ will be continuos at $x=x_o$ if $\lim_{x \to x_o^-} ƒ(x)=\lim_{x \to x_o^+} ƒ(x)=ƒ(x_o)$ If any of the above conditions is not satisfied then the function $ƒ(x)$ is said to be discontinuous at that point.

## Types of Discontinuities

A discontinuous function may be of the following types:

1. If limit of the function $ƒ(x)$ does not exist i.e. $\lim_{x \to x_o^-} ƒ(x) ≠ \lim_{x \to x_o^+} ƒ(x)$ then $ƒ(x)$ is said to be an ordinary discontinuity or a jump.
2. If $\lim_{x \to x_o} ƒ(x) ≠ ƒ(x_o)$ then the function $ƒ(x)$ is said to have a removable discontinuity at $x=x_o$. This type of discontinuity can be removed by redefining the function.
3. If $\lim_{x \to x_o} ƒ(x) \to \infty \text{ or } -\infty$ then $ƒ(x)$ is said to have an infinite discontinuity at $x=x_o$.

### Examples

The function $y=ƒ(x)=2x-3$ is continuos at every point.

A function $ƒ(x)$ defined by $ƒ(x)=\left\{ \begin{array}{c} 1 & \text{ for } x≥0 \\ 2 & \text{ for } x<0 \end{array}\right.$ is discontinuous at $x=0$. It is an ordinary discontinuity i.e. there is a jump at $x=0$.

A function $ƒ(x)$ defined by $y=ƒ(x)=\frac{x^2-4}{x-2}$ is discontinuous at $x=2$. It is because $ƒ(2)$ does not exist since $ƒ(2)=\frac{0}{0}$ which is an indeterminate form.

A function $ƒ(x)$ defined by $y=ƒ(x)=\frac{1}{x-2}$ is discontinuous as $\lim_{x \to 2^-} \frac{1}{x-2} \to -\infty$ $\lim_{x \to 2^+} \frac{1}{x-2} \to \infty$. It is an infinite discontinuity.

# Continuity in an Interval

A function $ƒ(x)$ is said to be continuos in an open interval $(a, b)$, if it is continuos at every point in $(a, b)$.

A function $ƒ(x)$ is said to be continuos in a closed interval $[a, b]$, if it is continuos at every point of the open interval $(a, b)$ and if it is continuos at the point $a$ from the right and continuos at the point $b$ from the left i.e. $\lim_{x \to a^+} ƒ(x) = ƒ(a) \text{ and } \lim_{x \to b^-} ƒ(x)=ƒ(b)$

Q. A function $ƒ(x)$ is defined as follows: $ƒ(x)=\left\{ \begin{array}{c} 2x-3 & \text{ for } x<2 \\ 2 & \text{ for } x=2 \\ 3x-5 & \text{ for } x>2 \end{array} \right.$ Is the function $ƒ(x)$ continuos at $x=2$? If not, how can the function $ƒ(x)$ be made continuos at $x=2$?

Left Hand Limit at $x=2$ is $\lim_{x \to 2^-} (2x-3)=2×2-3=1$ Right Hand Limit at $x=2$ is $\lim_{x \to 2^+} (3x-5)=3×2-5=1$ Since LHL and RHL are finite and equal, $\lim_{x \to 2} ƒ(x) \text{ exists and } \lim_{x \to 2} ƒ(x)=1$ But $ƒ(2)=2$. $\therefore \lim_{x \to 2} ƒ(x) ≠ ƒ(2)$ Hence, ƒ(x) is not continuos at $x=2$. This is a case of removable discontinuity. The given function will be continuos if $ƒ(2)=1$. Thus, the given function can be made continuos by defining the function in the following way: $ƒ(x)=\left\{ \begin{array}{c} 2x-3 & \text{ for } x<2 \\ 1 & \text{ for } x=2 \\ 3x-5 & \text{ for } x>2 \end{array} \right.$

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