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Q. If $f(x)=\begin{cases}1-2 x ; & x<0 \\ 2 & x=0 \\ x^{2}+2 ; & x>0\end{cases}$ then at $x=0$

KCETKCET 2022

Solution:

At $x =0 LHL =\lim\limits _{x \rightarrow 0-} 1-2 x =1 ; RHL =\lim \limits_{x \rightarrow 0+} x^{2}+2=2 ; f (0)=2 ; LHL \# RHL = f ( o )$
$f$ is continuous from right but discontinuous from left