Question

Evaluate the left hand and right hand limits of the function defined by $f\left(x\right)=\{\begin{array}{cc}1+x^2,& \text{if}0\le x\le 1\\ 2-x,& \text{if}x>1\end{array}$ at $x=1$. Also show that $\underset{x\to 1}{lim}f\left(x\right)$ does not exist.

Answer 1

Left hand limit:

$\underset{x\to 1}{lt}f\left(x\right)$

$=\underset{x\to 1}{lt}1+x^2$ $[\because f(x)=1+x^2$ when $x\le 1]$

$=1+1$

$=2$

Right hand limit

$\underset{x\to 1^{+}}{lt}f\left(x\right)$

$=\underset{x\to 1}{lt}2-x$ $[\because f(x)=2-x$ when $x>1]$

$=1$

Since, $\underset{x\to 1^{-}}{lt}f\left(x\right)\ne \underset{x\to 1^{+}}{lt}f\left(x\right),\underset{x\to 1}{lt}f\left(x\right)$ does not exist.