Continuity and Differentiability
1.0 Continuous Function
1.0 Continuous Function
Definition:
A function is continuous at $x=c$ if the function is defined at $x=c$ and the value at $x=c$ equals the left hand limit and right hand limit of the function at $x=c$ i.e., $$\mathop {\lim }\limits_{x \to c} f(x) = exists$$and$$\mathop {\lim }\limits_{x \to {c^ + }} f(x) = \mathop {\lim }\limits_{x \to {c^ - }} f(x) = \mathop {\lim }\limits_{x \to c} f(x)$$
Explanation:
Let us consider a simple function \[f(x) = \left\{ {\begin{array}{c}2&{if{\text{ }}x \leqslant 0} \\ 3&{if{\text{ }}x > 0} \end{array}} \right.\]First of all, we plot this function as shown. We get two straight lines $(1)$ and $(2)$ parallel to $X-$axis.
From the graph, it is clear that for all the values of ${x \leqslant 0}$, value of function is $$f(x)=2$$For all the values of $x>0$, value of function is $$f(x)=3$$$\therefore$ We can say that at $x=0$, $${\text{Left hand limit }} \ne {\text{ Right hand limit}}$$$\therefore$The function is not a continuous function at $x=0$.
- In simple words, we can not draw the graph of the function without lifting the pen which creates a discontinuity at $x=0$. Hence, the above function is not a continuous function at $x=0$.
Question 1. Explain the continuity of the function $f$ given by $f(x) = \left| x \right|$ at $x=0$.
Solution: We can write the modulus function as \[f(x) = \left\{ {\begin{array}{c}{ - x}&{x < 0} \\ x&{x \geqslant 0} \end{array}} \right.\]It is clearly shown that the function is defined at $x=0$ and $f(0)=0$.
Left hand limit of the function $f$ at $x=0$ is $$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} ( - x) = 0$$Similarly, Right hand limit of the function $f$ at $x=0$ is $$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (x) = 0$$Therefore, the left hand limit, right hand limit and the value of the function coincide at $x = 0$. Hence, $f$ is continuous at $x = 0$.
