Let \[f\left( x \right) = \left\{ \begin{array}{l}x + 1, & if x \geq 0 \\ x - 1, & if x < 0\end{array} . \right.\]Prove that \[\lim_{x \to 0} f\left( x \right)\] does not exist.

#### Solution

\[f\left( x \right) = \begin{cases}x + 1, & x \geq 0 \\ x - 1, & x < 0\end{cases}\]

\[\text{ RHL }: \]

\[ \lim_{x \to 0^+} f\left( x \right)\]

\[ = \lim_{x \to 0} \left( x + 1 \right)\]

\[\text{ Let } x = 0 + h, \text{ where } h \to 0 . \]

\[ \lim_{h \to 0} \left( 0 + h + 1 \right)\]

\[ = 1\]

\[\text{ LHL }: \]

\[ \lim_{x \to 0^-} f\left( x \right)\]

\[ = \lim_{x \to 0^-} \left( x - 1 \right)\]

\[\text{ Let } x = 0 - h, \text{ where } h \to 0 . \]

\[ \lim_{h \to 0} \left( 0 - h - 1 \right)\]

\[ = - 1\]

\[ LHL \neq RHL\]

\[\text{ Thus }, \lim_{x \to 0} f\left( x \right) \text{ does not exist } .\]