Limits to the Limit -3!

Calculus Level 3

lim ⁡ x → 0 ( 1 − [ cos ⁡ ( x ) cos ⁡ ( 2 x ) ] tan ⁡ 2 x ) \large \lim_{x\to 0} \left( \frac{1 - \left[\cos(x) \sqrt{\cos(2x)} \right]}{\tan^2 x} \right)

If the limit above can be expressed as A B \frac AB where A , B A,B are coprime positive integers, find A + B A+B .


The answer is 5.

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Aditya Tiwari by Aditya Tiwari , 22, India

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1 solution

Brian Lie
Mar 19, 2018

L = lim ⁡ x → 0 1 − cos ⁡ x cos ⁡ 2 x tan ⁡ 2 x = lim ⁡ x → 0 1 − cos ⁡ x tan ⁡ 2 x + lim ⁡ x → 0 ( 1 − cos ⁡ 2 x ) cos ⁡ x tan ⁡ 2 x = lim ⁡ x → 0 1 − cos ⁡ x x 2 + lim ⁡ x → 0 1 − cos ⁡ 2 x x 2 ( 1 + cos ⁡ 2 x ) = lim ⁡ x → 0 1 2 x 2 x 2 + lim ⁡ x → 0 1 2 ( 2 x ) 2 2 x 2 = 3 2 \begin{aligned} L&=\lim_{x\to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{\tan^2 x} \\&=\lim_{x\to 0}\frac{1-\cos x}{\tan^2 x}+\lim_{x\to 0}\frac{(1-\sqrt{\cos 2x})\cos x}{\tan^2 x} \\&=\lim_{x\to 0}\frac {1-\cos x}{x^2}+\lim_{x\to 0}\frac {1-\cos 2x}{x^2(1+\sqrt{\cos 2x})} \\&=\lim_{x\to 0}\frac{\frac 12x^2}{x^2}+\lim_{x\to 0}\frac {\frac 12 (2x)^2}{2x^2} \\&=\frac 32 \end{aligned} Therefore, the answer is 3 + 2 = 5 3+2=\boxed 5 .

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ohh!!!! i thought it was GIF in square bracket btw nice solution

Ashutosh Sharma - 3 years, 2 months ago

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