Let f: R → R be defined as f(x) = \ cases x ^ 3 ( 1 - 2x ) ^ 2 _ e...

A function $f\left(x\right)$ is continuous at a point $x = a,$ if $\lim_{x \to a}f\left(x\right) = f\left(a\right) .$

Thus, for continuity of the given function, we have

$\lim_{x \to 0}f\left(x\right) = \alpha$

$⇒ \lim_{x \to 0}\left[\dfrac{{x}^{3}}{( 1 - \cos 2x{)}^{2}}\log_{e}\left(\dfrac{1 + 2x{e}^{- 2x}}{{\left(1 - x{e}^{- x}\right)}^{2}}\right)\right] = \alpha$

Using, $\cos 2x = 1 - 2\sin^{2}x, \log_{e}\left(\dfrac{m}{n}\right) = \log_{e}m - \log_{e}n$ $\text{ and } \log_{e}{m}^{n} = n{\log}_{e}m,$

we get

$\lim_{x \to 0}\left[\dfrac{{x}^{3}}{{\left(2\sin^{2}x\right)}^{2}}\left(\log_{e}\left(1 + 2x{e}^{- 2x}\right) - \log_{e}{\left(1 - x{e}^{- x}\right)}^{2}\right)\right] = \alpha$

$⇒ \lim_{x \to 0}\left[\dfrac{{x}^{3}}{4\sin^{4}x}\left(\log_{e}\left(1 + 2x{e}^{- 2x}\right) - 2\log_{e}\left(1 - x{e}^{- x}\right)\right)\right] = \alpha$

$⇒ \lim_{x \to 0}\left[\dfrac{{x}^{4}}{4x\sin^{4}x}\left(\dfrac{2x{e}^{- 2x} \cdot \log_{e}\left(1 + 2x{e}^{- 2x}\right)}{2x{e}^{- 2x}} + \dfrac{2x{e}^{- x} \cdot \log_{e}\left(1 - x{e}^{- x}\right)}{- x{e}^{- x}}\right)\right] = \alpha$

Now, using the standard limits $\lim_{x \to 0}\left(\dfrac{\sin x}{x}\right) = 1 \text{ and } \lim_{x \to 0}\left(\dfrac{\log_{e}\left(1 + x\right)}{x}\right) = 1,$ we get

$⇒ \lim_{x \to 0}\left[\dfrac{1}{4x}\left(2x{e}^{- 2x} + 2x{e}^{- x}\right)\right] = \alpha$$⇒ \lim_{x \to 0}\left[\dfrac{2x}{4x}\left({e}^{- 2x} + {e}^{- x}\right)\right] = \alpha$

$⇒ \lim_{x \to 0}\left[\dfrac{1}{2}\left({e}^{- 2x} + {e}^{- x}\right)\right] = \alpha$

$⇒ \dfrac{1}{2}\left(2\right) = \alpha$

$⇒ \alpha = 1 .$