Continuity | Limits and Continuity

Continuity of a function means whether the function is continuous or discontinuous. 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

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

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:

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.
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.
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 of Continuity

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 $x=0$.

Graph of a discontinuous function showing ordinary discontinuity or jump

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.

Graph showing discontinuity in a function because of indeterminate form 0/0

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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