Question

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

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{c}\begin{array}{c}1+x^2,\mathrm{if}0\le \mathrm{x}\le 1\\ 2-x,\mathrm{if}x>1\end{array}\end{array}\right.$
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We observe
(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 }\left(1+{\left(1-h\right)}^2\right)=\underset{h\to 0}{\lim }\left(2+h^2-2h\right)=2$


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


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

Thus, f(x) is discontinuous at x = 1.