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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

\lim\limits_{x\to 0}\frac{\cos 4x-4\cos 2x+3}{x^{4}}=

4
8
$\frac{1}{4}$
$\frac{1}{8}$

Solution:

Consider the limit

\lim\limits_{x\to 0}\frac{\cos 4x-4\cos 2x+3}{x^{4}}

The above limit is solved as,

\lim\limits_{x\to 0}\frac{\cos 4x-4\cos 2x+3}{x^{4}}

=\lim\limits_{x\to 0}\frac{-4\sin 4x+8\sin 2x}{4x^{3}}

=\lim\limits_{x\to 0}\frac{-16\cos 4x+16\cos 2x}{12x^{2}}

=\lim\limits_{x\to 0}\frac{64\sin 4x-32\sin 2x}{24x}

=\lim\limits_{x\to 0}\frac{256\cos 4x-64\cos 2x}{24}

Solve further,

\lim\limits_{x\to 0}\frac{\cos 4x-4\cos 2x+3}{x^{4}}

=\frac{256-64}{24}

=\frac{192}{24}

=8

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