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Question
Find the value of 'a' for which the function f defined by
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Solution
\[f\left( x \right) = \binom{a \sin \frac{\pi}{2}\left( x + 1 \right), x \leq 0}{\frac{\tan x - \sin x}{x^3}, x > 0}\]
We have
(LHL at x = 0) =
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} a \sin \frac{\pi}{2}\left( - h + 1 \right) = a \sin\frac{\pi}{2} = a\]
(RHL at x = 0) =
\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \frac{\tan h - \sin h}{h^3}\]
\[\Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h} - \sin h}{h^3}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h}\left( 1 - \cos h \right)}{h^3}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\left( 1 - \cos h \right)\tan h}{h^3}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}\tan h}{4\frac{h^2}{4} \times h}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{2}{4} \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}\tan h}{\frac{h^2}{4} \times h}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^2 \lim_{h \to 0} \frac{\tan h}{h}\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2} \times 1 \times 1\]
\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2}\]
\[If f\left( x \right) \text{is continuous at} x = 0, then\]
\[ \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[ \Rightarrow a = \frac{1}{2}\]
