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CGPET 2016 Solved Paper

Section: Mathematics

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Question : 114 of 150

Marks: +1, -0

The value of

\lim\limits_{x \to 0}\;\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}

0
$-\frac{1}{6}$
$\frac{1}{6}$
Does not exist

Solution:

\lim\limits_{x \to 0}\;\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}

Using L'Hospital Rule

=\lim\limits_{x \to 0}\frac{4x^7\sin x^4+20x^{19}}{8e^{2x^4}x^7-16x^7+4e^{2x^4}-4x^3}

=\lim\limits_{x \to 0}\frac{4x^3(x^4\sin x^4+5x^{16})}{4x^3(2e^{2x^4}x^4-4x^4+e^{2x^4}-1)}

Using L'Hospital Rule

=\lim\limits_{x \to 0}\frac{4x^3\sin^4+4x^7\cos x^4+80x^{15}}{16e^{2x^4}x^7+16e^{2x^4}x^3-16x^3}

=\lim\limits_{x \to 0}\frac{4x^3(\sin x^4+x^4\cos x^4+20x^{12})}{4x^3(4e^{2x^4}+4e^{2x^4}-4)}

Using L'Hospital Rule

=\lim\limits_{x \to 0}\frac{4x^3\cos x^4+4x^3\cos x^4-4x^7\sin x^4+240x^{11}}{32e^{2x^4}\cdot x^7+48e^{2x^4}\cdot x^3}

=\lim\limits_{x \to 0}\frac{4x^3(2\cos x^4-x^4\sin x^4+60x^{11})}{4x^3(8e^{2x^4}\cdot x^4+12e^{2x^4})}

=\lim\limits_{x \to 0}\frac{2\cos x^4-x^4\sin x^4+60x^{11}}{e^{2x^4}(8x^4+12)}

=\lim\limits_{x \to 0}\frac{2\cos(0)^4-(0)^4\sin(0)^4+60(0)^{11}}{e^{2(0)^2}[8(0)^4+12]}

=\frac{2}{12}=\frac{1}{6}

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