Question

Find all points of discontinuity of $f$, where $f$ is
defined by $f\left(x\right)=\{\begin{array}{c}x+1ifx\ge 1\\ x^2+1ifx<1\end{array}$

Answer 1

Case $I$

At $x=1$

$f$ is continuous at $x=1$

L.H.L$=$R.H.L$=f\left(1\right)$

L.H.L$={lim}_{x\to 1^{-}}f\left(x\right)$

$={lim}_{x\to 1^{-}}(x^2+1)$

$=1+1=2$

R.H.L$={lim}_{x\to 1^{+}}f\left(x\right)$

$={lim}_{x\to 1^{+}}(x+1)$

$=1+1=2$

$f\left(x\right)=x+1$

$f\left(1\right)=1+1=2$

Thus, L.H.L$=$R.H.L$=f\left(1\right)$

$f$ is continuous at $x=1$

Case$II$

Let $x=c$ where $c<1$

$\therefore f\left(x\right)=x^2+1$ as $x=c$ and $c<1$

$f$ is continuous at $x=c$ iff ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

L.H.S$={lim}_{x\to c}f\left(x\right)$

$={lim}_{x\to c}(x^2+1)$

$=c^2+1$

R.H.S$={lim}_{x\to c}f\left(x\right)$

$={lim}_{x\to c}(x^2+1)$

$=c^2+1$

Thus,${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

$\Rightarrow f$ is continuous for $x=c$ where $c<1$

$\Rightarrow f$ is continuous for all real numbers less than $1$

Case$III$

Let $x=c$ where $c>1$

$f\left(x\right)=x+1$ as $x=c,c>1$

$f$ is continuous at $x=c$ if ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

L.H.S$={lim}_{x\to c}f\left(x\right)$

$={lim}_{x\to c}(x+1)=c+1$

R.H.S$=f\left(x\right)=x+1$

$f\left(c\right)=c+1$

Thus, ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

$\Rightarrow f$ is continuous for $x=c$ where $c<1$

$\Rightarrow f$ is continuous for all real numbers less than $1$

Hence, there is no point of discontinuity.

$\Rightarrow f$ is continuous at all real point.

Thus, $f$ is continuous for all $x\in R$