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Solve: \(\lim\limits_{x \to0}\frac{sin4x}{sin2x}\) |
Answer» = \(\frac{\lim\limits_{x \to 0}\frac{sin4x}{4x} . 4x}{\lim\limits_{x\to0}\frac{sin2x}{2x}.2x}\) = \(\frac{1\times 4}{1 \times2} = 2\) |
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