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प्रश्न
Show that
\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{\tan 2x} , if x < 0 \\ \frac{3}{2} , if x = 0 \\ \frac{\log(1 + 3x)}{e^{2x} - 1} , if x > 0\end{cases}\text{is continuous at} x = 0\]
उत्तर
Given:
\[f\left( x \right) = \begin{cases}\frac{\sin3x}{\tan2x}, if x < 0 \\ \frac{3}{2}, if x = 0 \\ \frac{\log\left( 1 + 3x \right)}{e^{2x} - 1}, if x > 0\end{cases}\]
We observe
(LHL at x = 0) =
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]
\[= \lim_{h \to 0} \left( \frac{\sin3\left( - h \right)}{\tan2\left( - h \right)} \right) = \lim_{h \to 0} \left( \frac{\sin3h}{\tan2h} \right) = \lim_{h \to 0} \left( \frac{\frac{3\sin3h}{3h}}{\frac{2\tan2h}{2h}} \right)\]
\[ = \frac{\lim_{h \to 0} \left( \frac{3\sin3h}{3h} \right)}{\lim_{h \to 0} \left( \frac{2\tan2h}{2h} \right)} = \frac{3 \lim_{h \to 0} \left( \frac{\sin3h}{3h} \right)}{2 \lim_{h \to 0} \left( \frac{\tan2h}{2h} \right)} = \frac{3 \times 1}{2 \times 1} = \frac{3}{2}\]
(RHL at x = 1) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right)\]
\[= \lim_{h \to 0} \left( \frac{\log\left( 1 + 3h \right)}{e^{2h} - 1} \right) = \lim_{h \to 0} \left( \frac{3h\frac{\log\left( 1 + 3h \right)}{3h}}{\frac{2h\left( e^{2h} - 1 \right)}{2h}} \right)\]
\[ = \frac{3}{2} \lim_{h \to 0} \left( \frac{\frac{\log\left( 1 + 3h \right)}{3h}}{\frac{\left( e^{2h} - 1 \right)}{2h}} \right) = \frac{3}{2}\frac{\lim_{h \to 0} \left( \frac{\log\left( 1 + 3h \right)}{3h} \right)}{\lim_{h \to 0} \left( \frac{\left( e^{2h} - 1 \right)}{2h} \right)} = \frac{3 \times 1}{2 \times 1} = \frac{3}{2}\]
And
\[f\left( 0 \right) = \frac{3}{2}\]
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
Thus, f(x) is continuous at x = 0.
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