limx→0 xtan⁡2x−2xtan⁡x(1−cos⁡2x)2, is

$\underset{x\to 0}{lim} \frac{x\mathrm{tan}2x-2x\mathrm{tan}x}{\left(1-\mathrm{cos}2x{\right)}^{2}}$, is

1. A

2

2. B

-2

3. C

$\frac{1}{2}$

4. D

$\frac{-1}{2}$

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Solution:

$\underset{x\to 0}{lim} \frac{x\mathrm{tan}2x-2x\mathrm{tan}x}{\left(1-\mathrm{cos}2x{\right)}^{2}}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{x\left(\mathrm{tan}2x-2\mathrm{tan}x\right)}{\left(1-\mathrm{cos}2x{\right)}^{2}}\\ =\underset{x\to 0}{lim} \frac{2x{\mathrm{tan}}^{3}x}{\left(1-{\mathrm{tan}}^{2}x\right)\left(1-\mathrm{cos}2x{\right)}^{2}}\\ =\underset{x\to 0}{lim} \frac{2{\left(\frac{\mathrm{tan}x}{x}\right)}^{3}}{\left(1-{\mathrm{tan}}^{2}x\right){\left(\frac{1-\mathrm{cos}2x}{{x}^{2}}\right)}^{2}}\\ =\frac{2×{1}^{3}}{\left(1-0\right)×4}=\frac{1}{2}\end{array}$

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