Limits with cosine

Calculus Level 3

$\lim_{x\to0} \dfrac{1 - \cos x \cos 2x \cos 3x}{\sin^2 2x } = \, ?$

\frac{7}{4}

\frac{3}{4}

\frac{8}{5}

1 0

\frac{2}{3}

2 solutions

Tom Van Lier
Mar 28, 2016

We can rewrite the numerator as:

$1 - cos(x) cos(2x) cos(3x) =$ $1 - \frac{1}{2} cos(2x) . (cos(4x) + cos(2x)) =$ $\frac{1}{2} . [1 + 1 - cos^2(2x) - cos(2x).cos(4x)] =$ $\frac{1}{2} . [ sin^2(2x) + 1 - cos^3(2x) + cos(2x) sin^2(2x)] =$ $\frac{1}{2} . [ sin^2(2x) + cos(2x) sin^2(2x) + cos(2x).sin^2(2x) + 1 - cos(2x) ] =$ $\frac{1}{2} . [ sin^2(2x) + 2.cos(2x).sin^2(2x) + 2.sin^2(x) ] .$ This means :

$\lim_{x \to 0} \dfrac{1 - cos(x) cos(2x) cos(3x) }{sin^2(2x)} =$ $\frac{1}{2} . \bigg[ \lim_{x \to 0} (1) + 2.\lim_{x \to 0} (cos(2x)) + 2 . \lim_{x \to 0} \dfrac{1}{4cos^2(x)} \bigg] =$ $\frac{1}{2} + 1 + \frac{1}{4} = \frac{7}{4}.$

FYI you used the term "nominator" instead of "numerator". I've edited the solution accordingly.

Calvin Lin Staff - 5 years, 2 months ago

Ok thanks, English isn't my native language :-)

Tom Van Lier - 5 years, 2 months ago

D S
Jan 15, 2018

We see sin^2(2x)= \frac{1-\cos{4x}}{2}

sub this in then limit falls apart using taylor series

Limits with cosine

2 solutions

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