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$\mathop {\lim }\limits_{x \to 0} \frac{{\log \cos x}}{x} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = $

$\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x} = $

If $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ and $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$,then

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}$ is equal to

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