Evaluate \[\lim_{x \to 2} f\left( x \right)\] (if it exists), where \[f\left( x \right) = \left\{ \begin{array}{l}x - \left[ x \right], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2\end{array} . \right.\]
Solution
\[f\left( x \right) = \begin{cases}x - \left[ x \right], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2\end{cases}\]
\[\text{ LHL }: \]
\[ \lim_{x \to 2^-} f\left( x \right)\]
\[ = \lim_{x \to 2^-} \left\{ x - \left[ x \right] \right\}\]
\[\text{ Let } x = 2 - \text{ h, where h } \to 0 . \]
\[ \Rightarrow \lim_{h \to 0} \left[ \left( 2 - h \right) - \left[ 2 - h \right] \right]\]
\[ = 2 - \left( 1 \right)\]
\[ = 1\]
\[\text{ RHL }: \]
\[ \lim_{x \to 2^+} f\left( x \right)\]
\[ = \lim_{x \to 2^+} \left( 3x - 5 \right)\]
\[\text{ Let } x = 2 + \text{ h, where h } \to 0 . \]
\[ \Rightarrow \lim_{h \to 0} \left[ 3\left( 2 + h \right) - 5 \right]\]
\[ = 6 - 5\]
\[ = 1\]
\[\text{ Here }, LHL = RHL = 1\]
\[ \therefore \lim_{x \to 2} f\left( x \right) = 1\]
