limx→0sin2x2xsin2x2x−sin2x=

limx→0sin2x2xsin2x2x−sin2x=elimx→0sin2x2x−sin2xsin2x2x−1 using limx→af(x) g(x)=elimx→ag(x) f(x)-1=elimx→0−sin2x2x=e−1=1e

limx→0sin2x2xsin2x2x−sin2x=

limx→0sin2x2xsin2x2x−sin2x=elimx→0sin2x2x−sin2xsin2x2x−1 using limx→af(x) g(x)=elimx→ag(x) f(x)-1=elimx→0−sin2x2x=e−1=1e

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$\lim_{x \to 0} {(\frac{\sin 2 x}{2 x})}^{(\frac{\sin 2 x}{2 x - \sin 2 x})} =$

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$\frac{1}{e}$

$e^{2}$

$e^{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \lim_{x \to 0} {(\frac{\sin 2 x}{2 x})}^{(\frac{\sin 2 x}{2 x - \sin 2 x})} \\ = e^{\lim_{x \to 0} (\frac{\sin 2 x}{2 x - \sin 2 x}) (\frac{\sin 2 x}{2 x} - 1)} u \sin g \lim_{x \to a} [f (x)]^{g (x)} = e^{\lim_{x \to a} g (x) (f (x) - 1)} \\ = e^{\lim_{x \to 0} (\frac{- \sin 2 x}{2 x})} \\ = e^{- 1} \\ = \frac{1}{e} \end{array}$