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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 62 of 160

Marks: +1, -0

\lim\limits_{x\to 0}

\frac{x^2(\tan 2x-2\tan x)^2}{(1-\cos 2x)^4}

Solution:

Let's consider the given expression as

P=\lim\limits_{x\to\infty}\;\frac{x^{2}(\tan 2x-2\tan x)^{2}}{(1-\cos 2x)^{4}}

Simplify the above expression

P=\lim\limits_{x\to\infty}\;\frac{x^2\left[\left(2x+\frac{(2x)^3}{3}+\frac{2}{15}(2x)^5+\cdots\right)-2\left(x+\frac{x^3}{3}+\frac{2}{15}x^5+\cdots\right)\right]^2}{\left[1-\left(1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\cdots\right)\right]^{4}}

=\lim\limits_{x\to\infty}\;\frac{4x^{8}\left[\left(\frac{4}{3}-\frac{1}{3}\right)+\frac{2}{15}(16x^{2}-x^{2})+\cdots\right]^{2}}{16x^{8}\left[1+\frac{x^{2}}{3}+\cdots\right]^{4}}

=\;\frac{4}{16}

=\;\frac{1}{4}

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