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Question: \(\lim_{x \rightarrow 0}\frac{x\tan 2x - 2x\tan x}{\left( 1 - \cos 2x \right)^{2}}\) is...

lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\lim_{x \rightarrow 0}\frac{x\tan 2x - 2x\tan x}{\left( 1 - \cos 2x \right)^{2}} is

A

2

B

−2

C

½

D

−1/2

Answer

½

Explanation

Solution

lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2=lim⁡x→0x{2x+8x33+64x515+....}−2x{x+x33+2x515+....}4sin⁡4x\lim_{x \rightarrow 0}\frac{x\tan 2x - 2x\tan x}{\left( 1 - \cos 2x \right)^{2}} = \lim_{x \rightarrow 0}\frac{x\left\{ 2x + \frac{8x^{3}}{3} + \frac{64x^{5}}{15} + .... \right\} - 2x\left\{ x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + .... \right\}}{4\sin^{4}x}

= lim⁡x→0x4{83−23+termscontainingpositivepowersofx}4sin⁡4x=14.2=12\lim_{x \rightarrow 0}\frac{x^{4}\left\{ \frac{8}{3} - \frac{2}{3} + termscontainingpositivepowersofx \right\}}{4\sin^{4}x} = \frac{1}{4}.2 = \frac{1}{2}

Alternative, = lim⁡x→0xtan⁡2x−2xtan⁡x4−sin⁡4x\lim_{x \rightarrow 0}\frac{x\tan 2x - 2x\tan x}{4 - \sin^{4}x}

= lim⁡x→0x4sin⁡4x[2tan⁡x1−tan⁡2x−2tan⁡x]\lim_{x \rightarrow 0}\frac{x}{4\sin^{4}x}\left\lbrack \frac{2\tan x}{1 - \tan^{2}x} - 2\tan x \right\rbrack

= lim⁡x→02xtan⁡x4sin⁡4x[1−1+tan⁡2x1−tan⁡2x]\lim_{x \rightarrow 0}\frac{2x\tan x}{4\sin^{4}x}\left\lbrack \frac{1 - 1 + \tan^{2}x}{1 - \tan^{2}x} \right\rbrack

=lim⁡x→02xtan⁡3xsin⁡4x(1−tan⁡2x)\lim_{x \rightarrow 0}\frac{2x\tan^{3}x}{\sin^{4}x\left( 1 - \tan^{2}x \right)}= 12lim⁡x→0xsin⁡x.1cos⁡3x.11−tan⁡2x\frac{1}{2}\lim_{x \rightarrow 0}\frac{x}{\sin x}.\frac{1}{\cos^{3}x}.\frac{1}{1 - \tan^{2}x}

= 12.1.113.11−0=12\frac{1}{2}.1.\frac{1}{1^{3}}.\frac{1}{1 - 0} = \frac{1}{2}