Question

Let ${lim}_{x\to a}f\left(x\right)$ exists but it is not equal to f (a). Then f(x) is discontinuous at x$=$ a and a is called a removable discontinuity. If ${lim}_{x\to a^{-}}f\left(x\right)=land{lim}_{x\to a^{+}}f\left(x\right)=m$ exist but $l\ne m.$ Then a is called a jump discontinuity. If one of the limits (left hand limit or right hand limit ) does not exist, then a is called an infinite discontinuity.

Let f(x)$\{\begin{array}{cc}2\left|x\right|,& x\le -1\\ 2x,& -1\le x\le 0\\ x+1,& 0<x\le 1\\ 2& x>1\end{array}$ Then $f\left(x\right)$ at

• A. $x=-1$ is a removable discontinuity
• B. $x=0$ is a jump discontinuity
• C. $x=1$ is a removable discontinuity
• D. $x=-1$ is a jump discontinuity

Answer 1

Answer: (B), (D)

$x=0$ is a jump discontinuity
$x=-1$ is a jump discontinuity

$LHL={lim}_{x\to {-1}^{-}}f\left(x\right)={lim}_{x\to -1}2(-x)=2$
$RHL={lim}_{x\to {-1}^{+}}f\left(x\right)={lim}_{x\to -1}2x=-2$
$\because$ Both the limits exist but not equal
Hence, $x=-1$ is a jump discontinuity.
$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}2\left(x\right)=0$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0}1+x=1$
$\because$ Both the limits exist but not equal
Hence, $x=0$ is a jump discontinuity.
$LHL={lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1}(x+1)=2$
$RHL={lim}_{x\to 1^{+}}f\left(x\right)=2$
Also, $f\left(1\right)=1+1=2$
Here, $LHL=RHL=f\left(1\right)$
Hence at $x=1$, $f$ is continuous.