Question

Let f(x) be a function defined by $f\left(x\right)=(\begin{array}{c}4x-5,\mathrm{if}x\le 2\\ x-k,\mathrm{if}x>2\end{array}$ If

$\underset{x\to 2}{\lim }f\left(x\right)$
exists, then the value of k is

Answer 1

We have, $f\left(x\right)=(\begin{array}{c}4x-5,\mathrm{if}x\le 2\\ x-k,\mathrm{if}x>2\end{array}$
$\therefore \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2-h\right)=\underset{h\to 0}{\lim }4\left(2-h\right)-5=\underset{h\to 0}{\lim }3-4h=3\mathrm{and}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2+h\right)=\underset{h\to 0}{\lim }\left(2+h\right)-k=2-k\because \underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)\Rightarrow 3=2-k\mathrm{or},k=-1$