Question
Find the value of 'a' for which the function f defined as
$ \begin{matrix} & \text{a sin}\frac{\pi}{2}\text{(x + 1)}, & x\leq0 \\ \text{f(x)} \\ & \frac{\text{tan x - sin x}}{\text{x}^{3}}, & x<0 \\ \end{matrix}$

is continuous at X = 0.

Answer

L.H.L = $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}$ f(x) = a
f(0) = a.sin$\pi/\text{r}$ = a
R.H.L. = $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}$ $\frac{\text{tan x}}{\text{x}}\cdot\frac{\text{(1 - cos x)}}{\text{x}^{2}}$
= $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\frac{\tan \text{x}}{\text{x}}\cdot2\Bigg(\frac{\sin\text{x/2}}{2\text{x/2}}\Bigg)^{2}=\frac{1}{2}$
$\therefore\text{a}=\frac{1}{2}$

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