Question

limx=0
$\frac{sin2x}{x}$
​

Answer 1

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$\displaystyle \lim_{x \to 0} \frac{sin2x}{x}$

$= \displaystyle \lim_{x \to 0} \frac{sin2x}{2x} \times 2$

$= 2 \times \displaystyle \lim_{x \to 0} \frac{sin2x}{2x}$

$= 2 \times 1$

$= 2$

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Answer 2

Answer:

limxsin2x

= \displaystyle \lim_{x \to 0} \frac{sin2x}{2x} \times 2=x→0lim2xsin2x×2

= 2 \times \displaystyle \lim_{x \to 0} \frac{sin2x}{2x}=2×x→0lim2xsin2x

= 2 \times 1=2×1

= 2=2