Question

Let $\underset{x\to a}{lim}f\left(x\right)$ 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 $f\left(x\right)=\{\begin{array}{cc}{x}^2+\left|x\right|,& x<-5\\ 25& x=-5\\ \frac{3x^2}{5}+15& x>-5\end{array}$. Then $x=-5$ is

• A. a point of discontinuity
• B. a jump discontinuity
• C. a removable discontinuity
• D. an infinite discontinuity

Answer 1

The correct option is C a removable discontinuity

$LHL={lim}_{x\to -5^{-}}f\left(x\right)$
$={lim}_{x\to -5^{-}}(x^2+\left|x\right|)$
$={lim}_{x\to -5}(x^2-x)$
$LHL=25+5=30$
$RHL={lim}_{x\to -5^{+}}f\left(x\right)$
$={lim}_{x\to -5}(\frac{3x^2}{5}+15)$
$=\frac{3\times 25}{5}+15=30$
Here, $LHL=RHL\ne f(-5)$
$\therefore x=-5$ is a removable discontinuity