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Question
Let \[f\left( x \right) = \begin{cases}x + 5, & if x > 0 \\ x - 4, & if x < 0\end{cases}\] \[\lim_{x \to 0} f\left( x \right)\] does not exist.
Solution
We have,
\[f\left( x \right) = \begin{cases}x + 5, & if x > 0 \\ x - 4, & if x < 0\end{cases}\]
LHL of f(x) at x = 0
= \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} \left( 0 - h - 4 \right) = - 4\]
RHL of f(x) at x = 0
= \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} \left( 0 + h + 5 \right) = 5\]
Clearly,
\[\lim_{x \to 0^-} f\left( x \right) \neq \lim_{x \to 0^+} f\left( x \right)\]
Hence,
\[\lim_{x \to 0} f\left( x \right)\] does not exist.
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