Finite limit

Calculus Level 3

lim x 0 a sin x sin 2 x tan 3 x \lim_{x\to 0} \dfrac{a \sin x - \sin 2x}{\tan ^3x} The above limit has a finite value λ \lambda for some positive integer a a . Find the value of λ + a \lambda + a .

4 7 9 3

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1 solution

Zico Quintina
May 4, 2018

lim x 0 a sin x sin 2 x tan 3 x = lim x 0 a sin x 2 sin x cos x ( sin 3 x cos 3 x ) = lim x 0 cos 3 x sin x ( a 2 cos x ) sin 3 x = lim x 0 cos 3 x ( a 2 cos x ) sin 2 x = lim x 0 cos 3 x ( a 2 cos x ) 1 cos 2 x = lim x 0 cos 3 x ( a 2 cos x ) ( 1 cos x ) ( 1 + cos x ) [Now, letting a = 2 ] = lim x 0 cos 3 x ( 2 2 cos x ) ( 1 cos x ) ( 1 + cos x ) = lim x 0 2 cos 3 x ( 1 cos x ) ( 1 cos x ) ( 1 + cos x ) = lim x 0 2 cos 3 x ( 1 + cos x ) = 2 2 = 1 = λ \begin{aligned} \lim_{x\to0} \dfrac{a \sin x - \sin 2x}{\tan^3 x} &= \lim_{x\to0} \dfrac{a \sin x - 2 \sin x \cos x}{\left( \dfrac{\sin^3 x}{\cos^3 x} \right)} \\ \\ &= \lim_{x\to0} \dfrac{\cos^3 x \sin x (a - 2 \cos x)}{\sin^3 x} \\ \\ &= \lim_{x\to0} \dfrac{\cos^3 x (a - 2 \cos x)}{\sin^2 x} \\ \\ &= \lim_{x\to0} \dfrac{\cos^3 x (a - 2 \cos x)}{1 - \cos^2 x} \\ \\ &= \lim_{x\to0} \dfrac{\cos^3 x (a - 2 \cos x)}{(1 - \cos x)(1 + \cos x)} \qquad \qquad \small \text{[Now, letting } a=2] \\ \\ &= \lim_{x\to0} \dfrac{\cos^3 x (2 - 2 \cos x)}{(1 - \cos x)(1 + \cos x)} \\ \\ &= \lim_{x\to0} \dfrac{2 \cos^3 x (1 - \cos x)}{(1 - \cos x)(1 + \cos x)} \\ \\ &= \lim_{x\to0} \dfrac{2 \cos^3 x}{(1 + \cos x)} = \dfrac{2}{2} = 1 = \lambda \end{aligned} so the answer is λ + a = 1 + 2 = 3 \lambda + a = 1 + 2 = \boxed{3}

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