Question

For \[f(x)=\begin{cases}1,&x\leq0\\2,&x>0\end{cases}\], why is \[f\] discontinuous at \[x=0\]?

Options

• \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=2\].

• \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=1\].

• \[\lim_{x\to0^-}f(x)=1\] and \[\lim_{x\to0^+}f(x)=2\].

• \[f(0)\] is not defined.

Answer 1

The left-hand limit is \[1\], whereas the right-hand limit is \[2\]. Since LHL is not equal to RHL, the limit does not exist.