Question

For what value of k is the following function continuous at x = 2?
$f\left(x\right)=\left\{\begin{array}{cc}2x+1;& \mathrm{if}x<2\\ k;& x=2\\ 3x-1;& x>2\end{array}\right.$

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{c}2x+1,\mathrm{if}\mathrm{x}<2\\ k,x=2\\ 3x-1,x>2\end{array}\right.$

We have
(LHL at x = 2) = $\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\left(2-h\right)+1\right)=5$

(RHL at x = 2) = $\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2+h\right)$$=\underset{h\to 0}{\lim }3\left(2+h\right)-1=5$

Also, $f\left(2\right)=k$

If f(x) is continuous at x = 2, then
​$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right)$

$\Rightarrow 5=5=k$

Hence, for k = 5, $f\left(x\right)$ is continuous at $x=2$.