• $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$
• $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1+u)}{u} = 1$ (for $u \to 0$)

• $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
• $e^x = 1 + x + \frac{x^2}{2!} + \dots$
• $\log_e(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$

Step 1: Understand the condition for continuity. For the function $f(x)$ to be continuous at $x=0$, the limit of $f(x)$ as $x$ approaches $0$ must be equal to the value of the function at $x=0$, which is $\alpha$. Therefore, we need to evaluate: $\alpha = \mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{(\,1 - \cos 2x)^2} \log_e\left(\frac{1 + 2x e^{-2x}}{(1 - x e^{-x})^2}\right)$

Step 2: Simplify the denominator $(1 - \cos 2x)^2$. We use the double angle identity for cosine: $\cos 2x = 1 - 2\sin^2 x$. So, $1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x$. Then, $(1 - \cos 2x)^2 = (2\sin^2 x)^2 = 4\sin^4 x$. Substituting this back into the limit expression, we get: $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \log_e\left(\frac{1 + 2x e^{-2x}}{(1 - x e^{-x})^2}\right)$

Step 3: Simplify the logarithmic term. Using the property of logarithms $\log_e(\frac{A}{B^2}) = \log_e A - 2\log_e B$, we can rewrite the logarithmic part: $\log_e\left(\frac{1 + 2x e^{-2x}}{(1 - x e^{-x})^2}\right) = \log_e(1 + 2x e^{-2x}) - 2\log_e(1 - x e^{-x})$ The limit expression becomes: $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ \log_e(1 + 2x e^{-2x}) - 2\log_e(1 - x e^{-x}) \right]$

Step 4: Apply standard limits and Taylor expansions. We can rewrite the expression as: $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{1}{4} \left(\frac{x}{\sin x}\right)^4 \left[ \log_e(1 + 2x e^{-2x}) - 2\log_e(1 - x e^{-x}) \right]$ Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$, we have $\mathop {\lim }\limits_{x \to 0} \left(\frac{x}{\sin x}\right)^4 = 1^4 = 1$. Now, we focus on the limit of the logarithmic terms. As $x \to 0$, $e^{-2x} \to 1$ and $e^{-x} \to 1$. So, $2x e^{-2x} \to 0$ and $x e^{-x} \to 0$. We use the standard limit $\mathop {\lim }\limits_{u \to 0} \frac{\log_e(1+u)}{u} = 1$. For the first term: $\log_e(1 + 2x e^{-2x})$. Let $u = 2x e^{-2x}$. As $x \to 0$, $u \to 0$. $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1 + 2x e^{-2x})}{2x e^{-2x}} = 1$ So, $\mathop {\lim }\limits_{x \to 0} \log_e(1 + 2x e^{-2x}) = \mathop {\lim }\limits_{x \to 0} (2x e^{-2x}) = 0$. Similarly, for the second term: $\log_e(1 - x e^{-x})$. Let $v = -x e^{-x}$. As $x \to 0$, $v \to 0$. $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1 - x e^{-x})}{-x e^{-x}} = 1$ So, $\mathop {\lim }\limits_{x \to 0} \log_e(1 - x e^{-x}) = \mathop {\lim }\limits_{x \to 0} (-x e^{-x}) = 0$. This indicates that using the direct $\frac{\log(1+u)}{u}$ limit might not be sufficient if the arguments $u$ themselves go to zero. We need to consider higher order terms or a different approach.

Let's use Taylor expansions around $x=0$: $e^{-2x} = 1 - 2x + \frac{(-2x)^2}{2!} + \dots = 1 - 2x + 2x^2 + \dots$ $2x e^{-2x} = 2x(1 - 2x + \dots) = 2x - 4x^2 + \dots$ $\log_e(1 + (2x - 4x^2 + \dots)) = (2x - 4x^2 + \dots) - \frac{(2x - \dots)^2}{2} + \dots = 2x - 4x^2 - \frac{4x^2}{2} + \dots = 2x - 6x^2 + \dots$

$e^{-x} = 1 - x + \frac{(-x)^2}{2!} + \dots = 1 - x + \frac{x^2}{2} + \dots$ $x e^{-x} = x(1 - x + \dots) = x - x^2 + \dots$ $\log_e(1 - (x - x^2 + \dots)) = -(x - x^2 + \dots) - \frac{(-x + \dots)^2}{2} + \dots = -x + x^2 - \frac{x^2}{2} + \dots = -x + \frac{x^2}{2} + \dots$ So, $2\log_e(1 - x e^{-x}) = 2(-x + \frac{x^2}{2} + \dots) = -2x + x^2 + \dots$

Now substitute these back into the limit expression: $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ (2x - 6x^2 + \dots) - (-2x + x^2 + \dots) \right]$ $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ 4x - 7x^2 + \dots \right]$ We know $\sin x = x - \frac{x^3}{6} + \dots$, so $\sin^4 x = (x - \dots)^4 = x^4 - \dots$. $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4x^4} \left[ 4x + \dots \right]$ $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{1}{4x} \left[ 4x + \dots \right]$ $\alpha = \mathop {\lim }\limits_{x \to 0} \left( 1 + \dots \right) = 1$

Alternatively, using the standard limit $\mathop {\lim }\limits_{u \to 0} \frac{\log_e(1+u)}{u} = 1$: We rewrite the expression as: $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ \frac{\log_e(1 + 2x e^{-2x})}{2x e^{-2x}} (2x e^{-2x}) - 2 \frac{\log_e(1 - x e^{-x})}{-x e^{-x}} (-x e^{-x}) \right]$ As $x \to 0$, $e^{-2x} \to 1$ and $e^{-x} \to 1$. So, $2x e^{-2x} \to 0$ and $-x e^{-x} \to 0$. The terms $\frac{\log_e(1 + 2x e^{-2x})}{2x e^{-2x}}$ and $\frac{\log_e(1 - x e^{-x})}{-x e^{-x}}$ both tend to 1. $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ 1 \cdot (2x \cdot 1) - 2 \cdot 1 \cdot (-x \cdot 1) \right]$ $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} \left[ 2x + 2x \right]$ $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{x^3}{4\sin^4 x} (4x)$ $\alpha = \mathop {\lim }\limits_{x \to 0} \frac{4x^4}{4\sin^4 x}$ $\alpha = \mathop {\lim }\limits_{x \to 0} \left(\frac{x}{\sin x}\right)^4$ Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$, we have $\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} = 1$. Therefore, $\alpha = 1^4 = 1$

Step 5: Conclude the value of $\alpha$. Since $f$ is continuous at $x=0$, we have $\alpha = \mathop {\lim }\limits_{x \to 0} f(x)$. From Step 4, we found this limit to be 1. Thus, $\alpha = 1$.