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Q. $\begin{aligned} & -4,3 \\ & 6,3 \\ & -6,3 \\ & -4,9 \\ & 9,-4\end{aligned}$

KEAMKEAM 2021

Solution:

$=\displaystyle\lim _{x \rightarrow-2^{-}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(-2-h) $
$ =\displaystyle\lim _{h \rightarrow 0}[3(-2-h)+2] $
$ =\displaystyle\lim _{h \rightarrow 0}(-6-3 h+2) $
$ =-4 $
$\displaystyle\lim _{h \rightarrow-2^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(-2+h) $
$ =\displaystyle\lim _{h \rightarrow 0}\left[(-2+h)^2-3(-2+h)-1\right] $
$ =\displaystyle\lim _{h \rightarrow 0}\left(4+h^2-4 h+6-3 h-1\right) $
$ =\displaystyle\lim _{h \rightarrow 0}\left(9+h^2-7 h\right)=9$