Let f:RR be defined as f(x)=casesx^3(1-2x)^2_e(1+2xe^-2x(1-xe^-x)^2), & x0, & x=0cases.

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be defined as $f(x)=\begin{cases}\dfrac{x^3}{(1-\cos2x)^2}\log_e\left(\dfrac{1+2xe^{-2x}}{(1-xe^{-x})^2}ight), & x
e0\\alpha, & x=0\end{cases}$. If $f$ is continuous at $x=0$, then $\alpha$ is equal to

Accepted Answer

Using $\ln(1+2xe^{-2x})\approx2xe^{-2x}$ and $\ln(1-xe^{-x})^2\approx-2xe^{-x}$ for small $x$, and $1-\cos2x\approx2x^2$ so $(1-\cos2x)^2\approx4x^4\sin^4$-type behavior:
$$\alpha=\lim_{x\to0}\frac{x^3}{4x^4}\cdot2x(e^{-2x}+e^{-x})=\lim_{x\to0}\frac{2x(e^{-2x}+e^{-x})}{4x}=\frac{2\cdot2}{4}=1$$

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