Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2}{2},& \mathrm{if}0\le x\le 1\\ 2x^2-3x+\frac{3}{2},& \mathrm{if}1<x\le 2\end{array}\right.$. Show that f is continuous at x = 1.

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2}{2},\mathrm{if}0\le \mathrm{x}\le 1\\ 2x^2-3x+\frac{3}{2},\mathrm{if}1<\mathrm{x}\le 2\end{array}\right.$

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

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

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

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

Hence, the given function is continuous at $x=1$.