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(A) For $f(x)$ to be continuous at $x=0$,we must have $\alpha = \lim_{x \rightarrow 0} f(x)$.Given $f(x) = \frac{x^3}{(1-\cos 2x)^2} \ln\left(\frac{1+

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(A) For $f(x)$ to be continuous at $x=0$,we must have $\alpha = \lim_{x \rightarrow 0} f(x)$.Given $f(x) = \frac{x^3}{(1-\cos 2x)^2} \ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$.Since $1-\cos 2x = 2\sin^2 x$,we have $(1-\cos 2x)^2 = 4\sin^4 x$.So,$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{x^3}{4\sin^4 x} \ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$.Using $\sin x \approx x$ as $x \rightarrow 0$,$\frac{x^3}{4\sin^4 x} \approx \frac{x^3}{4x^4} = \frac{1}{4x}$.Now,$\ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right) = \ln(1+2xe^{-2x}) - 2\ln(1-xe^{-x})$.Using $\ln(1+u) \approx u$ for small $u$,we get $2xe^{-2x} - 2(-xe^{-x}) = 2xe^{-2x} + 2xe^{-x}$.As $x \rightarrow 0$,$e^{-2x} \rightarrow 1$ and $e^{-x} \rightarrow 1$,so the expression becomes $2x + 2x = 4x$.Thus,$\alpha = \lim_{x \rightarrow 0} \frac{1}{4x} \cdot (4x) = 1$.

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} \frac{x^{3}}{(1-\cos 2x)^{2}} \log_{e}\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^{2}}\right), & x \neq 0 \\ \alpha, & x=0 \end{cases}$. If $f$ is continuous at $x=0$,then $\alpha$ is equal to:

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