Chapter 14, Problem 12CT

To determine

T find: f is continuous or not at $x=-2,1,3,4$

Answer to Problem 12CT

(a) function is not continuous at $x=1$

(b)given function is continuous at

  $x=-2.$ (c)the function is not continuous at $x=3$

(d)at $x=4$ the function is continuous

Explanation of Solution

Given information:

  

(a) at $x=-2$

At first we find the right limit :

Here , $\underset{x\to -2^{+}}{\lim }f(x)=2$

Left limit

  $\underset{x\to -2^{-}}{\lim }f(x)=2$

[as same from left limit]

Hence,

  $\underset{x\to -2}{\lim }f(x)=2$

And function value at

  $x=-2$

Hence, given function is continuous at

  $x=-2$

(b)

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

So,

  $\underset{x\to 1}{\lim }f(x)=2$

And $f(1)=1$

Hence, $\underset{x\to 1}{\lim }f(x)\ne f(1)$

Hence, function is not continuous at $x=1$

(c)

  $\begin{array}{l}\underset{x\to 3^{+}}{\lim }f(x)=-3\\ \underset{x\to 3^{-}}{\lim }f(x)=5\end{array}$

Since,

  $\underset{x\to 3^{+}}{\lim }f(x)\ne \underset{x\to 3^{-}}{\lim }f(x)$

Limit do’t exist

So, the function is not continuous at $x=3$

(d)

  $\begin{array}{l}\underset{x\to 4^{+}}{\lim }f(x)=-4\\ \underset{x\to 4^{-}}{\lim }f(x)=-4\end{array}$

Limit exist and

  $f(4)=-4$

Hence, at $x=4$ the function is continuous.