Let $f : R \rightarrow R$ be defined as \[ f(x) = \begin{cases} \frac{x^3}{(1 - \cos 2x)^2} \cdot \log_e\left(\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}\right), & x \neq 0 \\ \alpha, & x = 0 \end{cases} \] . If $f$ is continuous at $x = 0$, then $\alpha$ is equal to:
Continuity Condition Analysis
For the function $f$ to be continuous at $x = 0$, the limit of the function as $x$ approaches $0$ must be equal to the function's value at $x = 0$. Mathematically, this means:
$ \lim_{x \to 0} f(x) = f(0) $
Given $f(0) = \alpha$, we need to compute the limit:
$ \alpha = \lim_{x \to 0} \frac{x^3}{(1 - \cos 2x)^2} \cdot \log_e\left(\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}\right) $
Limit Calculation Steps
We calculate the limit using standard calculus techniques and approximations:
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Simplify the denominator using trigonometric identities: We know $1 - \cos 2x = 2 \sin^2 x$. Therefore,
$ (1 - \cos 2x)^2 = (2 \sin^2 x)^2 = 4 \sin^4 x $
- Apply the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$: As $x \to 0$, $\sin x \approx x$. So, the denominator term $4 \sin^4 x$ behaves like $4x^4$.
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Analyze the logarithm term: Let $L = \log_e\left(\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}\right)$. As $x \to 0$, the argument $\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2} \to \frac{1+0}{(1-0)^2} = 1$.
We use the standard limit $\log_e(1+u) \approx u$ for small $u$. Let $u = \frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2} - 1$. We need to find the behavior of $u$ as $x \to 0$. $ u = \frac{1 + 2xe^{-2x} - (1 - xe^{-x})^2}{(1 - xe^{-x})^2} $ Using Taylor expansions $e^{-2x} \approx 1-2x$ and $e^{-x} \approx 1-x$: Numerator $\approx 1 + 2x(1-2x) - (1 - x(1-x))^2 = 1 + 2x - 4x^2 - (1 - x + x^2)^2$. Expanding $(1 - x + x^2)^2 \approx 1 + x^2 - 2x + 2x^2 = 1 - 2x + 3x^2$. Numerator $\approx 1 + 2x - 4x^2 - (1 - 2x + 3x^2) = 4x - 7x^2$. The denominator $(1 - xe^{-x})^2 \to 1^2 = 1$. So, $u \approx 4x$. Therefore, $L = \log_e(1+u) \approx u \approx 4x$.
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Combine terms and evaluate the limit: Substitute the approximations into the limit expression for $\alpha$:
$ \alpha = \lim_{x \to 0} \frac{x^3}{4 \sin^4 x} \cdot \log_e\left(\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}\right) $
$ \alpha \approx \lim_{x \to 0} \frac{x^3}{4x^4} \cdot (4x) = \lim_{x \to 0} \frac{4x^4}{4x^4} = 1 $
A more formal way using standard limits:
$ \alpha = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^4 \cdot \frac{1}{4} \cdot \log_e\left(\frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}\right) $
We need the limit of the logarithm term. Let $y = \frac{1 + 2xe^{-2x}}{(1 - xe^{-x})^2}$. $\lim_{x \to 0} \log_e y = \log_e 1 = 0$. Consider $\lim_{x \to 0} \frac{\log_e y}{y-1}$. This limit is 1. We found $y-1 \approx 4x$. So the original limit becomes:
$ \alpha = \lim_{x \to 0} \frac{x^3}{4 \sin^4 x} \cdot (4x) = \lim_{x \to 0} \frac{4x^4}{4 \sin^4 x} $
$ \alpha = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^4 = 1^4 = 1 $
Final Result
The calculation shows that the limit of the function $f(x)$ as $x$ approaches $0$ is $1$. For $f$ to be continuous at $x=0$, we must have $\alpha = \lim_{x \to 0} f(x)$. Therefore, $\alpha = 1$.
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