Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x^2}{2},if0\le x\le 1\\ 2x^2-3x+\frac{3}{2},if1<x\le 2\end{array}$ , Show that $f$ is continuous at $x=1$.

Answer 1

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2}{2},& \text{if}0\le x\le 1\\ 2x^2-3x+\frac{3}{2},& \text{if}1<x\le 2\end{array}$

LHL at $x=1$,

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}\frac{{(1-h)}^2}{2}=\frac{1}{2}$

RHL at $x=1$,

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}[2{(1+h)}^2-3(1+h)+\frac{3}{2}]=2-3+\frac{3}{2}=\frac{1}{2}$

Also,

$f\left(1\right)=\frac{1^2}{2}=\frac{1}{2}$

$\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$

Hence $f\left(x\right)$ is continuous at $x=1$.