Question

Find k so that ${lim}_{x\to 2}$ f(x) may exist, where $f\left(x\right)=\{\begin{array}{cc}2x+3,& x\le 2\\ x+k,& x>2\end{array}$

Answer 1

${lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{-}}(2x+3)$

=2(2)+3

=7

$\therefore {lim}_{x\to 2^{-}}f\left(x\right)=7$

Also,

${lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{+}}(x+k)$

=(2+k)

Since, ${lim}_{x\to 2}f\left(x\right)$ exists (given)

$\therefore {lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{+}}f\left(x\right)$

$\Rightarrow 7=2+k$

$\Rightarrow k=5$