question_answer

If \[f(x)=\left\{ \begin{align}   & \,\,\,\,\,\,\,x,\ \text{when }0\le x\le 1 \\  & 2-x,\ \text{when }1<x\le 2 \\ \end{align} \right.\],  then \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\]

A)                 1

B)                 2

C)                 0

D)                 Does not exist

Correct Answer: A

Solution :

                   Hence \[\underset{x\to 1}{\mathop{\lim }}\,\,f(x)=1\]                    Aliter :\[\underset{x\to 1-}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,(1-h)=1\]                    and \[\underset{x\to 1+}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,2-(1+h)=1\]                                 Hence limit of function is 1.