Chapter 2, Problem 208RE

wTrue or False. In the following exercises, justify your answer with a proof or a counterexample.

208. A function has to be continuous at x = a if the $\underset{x\to a}{\lim }f\left(x\right)$ exists.

To determine

To check: whether the statement, “A function has to be continous at $x=a$ if the $\underset{x\to a}{\lim }f\left(x\right)$ exists” is true or false.

Answer to Problem 208RE

The statement is false.

Explanation of Solution

Given information:

The given statement is, “A function has to be continous at $x=a$ if the $\underset{x\to a}{\lim }f\left(x\right)$ exists”.

Concept used:

The function $f(x)$ is said to be continuous at point $a$ if

  $f(a)$ is defined; $\underset{x\to a}{\lim }f(x)$ exist; and $\underset{x\to a}{\lim }f(x)=f(a)$ .

Calculation:

A counter example can be given as:

Consider the function,

  $f\left(x\right)=\left\{\begin{array}{l}{x}^3\text{+1, if }x<1\\ \text{3, if }x=1\\ \text{3}-x^2\text{, if }x>1\end{array}\right\}$

To classify continuity at $x=1$ we evaluate,

  $\begin{array}{l}\underset{x\to 1^{-}}{\lim }f\left(x\right)={\left(1\right)}^3+1=2\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=3-{\left(1\right)}^2=2\end{array}$

Thus, $\underset{x\to 1}{\lim }f\left(x\right)=2$ , but $f\left(1\right)=3$ .

So, $\underset{x\to 1}{\lim }f\left(x\right)\ne f\left(1\right)$

Therefore, $\underset{x\to 1}{\lim }f\left(x\right)$ exists but $f\left(x\right)$ is not continuous at $x=1$ .

Conclusion:

Hence, the given statement, “A function has to be continous at $x=a$ if the $\underset{x\to a}{\lim }f\left(x\right)$ exists” is false.