Question

Show that the function $f\left(x\right)=$ $\{\begin{array}{cc}3x-2& \text{if}0<x\le 1\\ 2x^2-x& \text{if}1<x\le 2\\ 5x-4& \text{if}x>2\end{array}$
is continuous at $x=2$

Answer 1

Left hand limit at $x=2$ is given as

${lim}_{x\to 2^{-}}f\left(x\right)=2x^2-x2(2{)}^2-22(4)-28-2=6$

Right hand limit at $x=2$ is given as

${lim}_{x\to 2^{+}}f\left(x\right)=5x-45\left(2\right)-410-46$

the value of $f\left(x\right)$ at $x=2$ is given as

$2x^2-22(2{)}^2-22(4)-28-2=6$

therfore, it is continous at $x=2$