Mathematics - Limits Question with Solution | TestHub

Since, $f(x)=\{\begin{array}{ll}{x}^2-1,& 0<x<2\\ 2x+3,& 2\le x<3\end{array}$
Now, ${\lim }_{x\to 2^{-}}f(x)={\lim }_{x\to 2^{-}}\left(x^2-1\right)$
$\begin{array}{rl}& =\underset{h\to 0}{\lim }\left[(2-h{)}^2-1\right]=\underset{h\to 0}{\lim }\left[4+h^2-4h-1\right]\\ & =\underset{h\to 0}{\lim }\left[h^2-4h+3\right]=0-0+3=3\end{array}$
and ${\lim }_{x\to 2^{+}}f(x)={\lim }_{x\to 2^{+}}(2x+3)$
$={\lim }_{h\to 0}[2(2+h)+3]={\lim }_{h\to 0}[4+2h+3]=7$
Hence the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7)x+(3+7)=0\Rightarrow x^2$ $-10x+21=0$