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प्रश्न
Let f(x) be a function defined by \[f\left( x \right) = \begin{cases}\frac{3x}{\left| x \right| + 2x}, & x \neq 0 \\ 0, & x = 0\end{cases} .\] Show that \[\lim_{x \to 0} f\left( x \right)\] does not exist.
उत्तर
\[f\left( x \right) = \begin{cases}\frac{3x}{\left| x \right| + 2x}, & x \neq 0 \\ 0, & x = 0\end{cases}\]
\[\text{ Left hand limit }: \]
\[ \lim_{x \to 0^-} \left[ \frac{3x}{\left| x \right| + 2x} \right]\]
\[\text{ Let } x = 0 - h, \text{ where } h \to 0 . \]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{3\left( - h \right)}{\left| - h \right| + 2\left( - h \right)} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{- 3h}{h - 2h} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{- 3h}{- h} \right]\]
\[ = 3\]
\[\text{ Right hand limit }: \]
\[ \lim_{x \to 0^+} \left( \frac{3x}{\left| x \right| + 2x} \right)\]
\[\text{ Let } x = 0 + h, \text{ where } h \to 0 . \]
\[ \Rightarrow^{} \lim_{h \to 0} \left( \frac{3h}{\left| h \right| + 2h} \right)\]
\[ = \lim_{h \to 0} \left( \frac{3h}{h + 2h} \right)\]
\[ = 1\]
\[ \lim_{x \to 0^-} \left( \frac{3x}{\left| x \right| + 2x} \right) \neq \lim_{x \to 0^+} \left( \frac{3x}{\left| x \right| + 2x} \right)\]
\[\text{ Thus }, \lim_{x \to 0} f\left( x \right) \text{ does not exist } .\]
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