Question

Find the value of 'a' for which the function f defined by

\[f\left( x \right) = \begin{cases}a\sin\frac{\pi}{2}(x + 1), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\]  is continuous at x = 0.

 

 

Answer 1

\[f\left( x \right) = \binom{a \sin \frac{\pi}{2}\left( x + 1 \right), x \leq 0}{\frac{\tan x - \sin x}{x^3}, x > 0}\]

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) = 

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \frac{\tan h - \sin h}{h^3}\]

\[\Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h} - \sin h}{h^3}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h}\left( 1 - \cos h \right)}{h^3}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{\left( 1 - \cos h \right)\tan h}{h^3}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}\tan h}{4\frac{h^2}{4} \times h}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{2}{4} \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}\tan h}{\frac{h^2}{4} \times h}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^2 \lim_{h \to 0} \frac{\tan h}{h}\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2} \times 1 \times 1\]

\[ \Rightarrow \lim_{x \to 0^+} f\left( x \right) = \frac{1}{2}\]

\[If f\left( x \right) \text{is continuous at} x = 0, then\]

\[ \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]

\[ \Rightarrow a = \frac{1}{2}\]