Question
Find k so that $\lim\limits_{\text{x}\rightarrow2}\text{f(x)}$ may exist, where $\text{f(x)}=\begin{cases}2\text{x}+3, & \text{x}\le 2\\\text{x}+\text{k}, & \text{x} > 2\end{cases}.$

Answer

$\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^-}(2\text{x}+3)$
$=2(2)+3$
$=7$
$\therefore\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=7$
Also,
$\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}(\text{x}+\text{k})$
$=(2+\text{k})$
Since, $\lim\limits_{\text{x}\rightarrow2}\text{f(x)}$ exists (given)
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}$
$\Rightarrow7=2+\text{k}$
$\Rightarrow\text{k}=5$

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