Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
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1$\begingroup$ Can you use Taylor series? $\endgroup$– HakimCommented Aug 5, 2014 at 12:40
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$\begingroup$ Compare with math.stackexchange.com/questions/87491/trigonometric-limit?rq=1. $\endgroup$– Dietrich BurdeCommented Aug 5, 2014 at 12:40
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$\begingroup$ @Hakim: bro no idea on taylor series... simplification of the trigonometric term will help... :( $\endgroup$– AdarshCommented Aug 5, 2014 at 12:47
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6 Answers
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Otherwise using: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ $$\eqalign{ \lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x} &=\lim_{x\to 0} \dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{2x\sin x\cos x}\\ &=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x\sin x\cos x}\\ &=\frac32\lim_{x\to 0} \dfrac{\frac{(1-\cos x)}x}{x\frac{\sin x}x\cos x}\\ &=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}\\ &=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}.\frac{1+\cos x}{1+\cos x}\\ &=\frac32\lim_{x\to 0} \dfrac{1}{\cos x(1+\cos x)}.\left(\frac{\sin x}{x}\right)^2\\ &=\frac32\times\frac1{1\times2}\times1^2=\frac34.}$$
Using Taylor series: $$\cos x =1-x^2/2+O(x^4)$$ $$\sin x =x-x^3/6+O(x^5)$$ $$(1+x)^n=1+nx+O(x^2)$$ So, $$\eqalign{ \lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x} &=\lim_{x\to 0} \dfrac{1-(1-x^2/2+O(x^4))^3}{x(2x+O(x^3))}\\ &=\lim_{x\to 0} \dfrac{1-(1-3x^2/2+O(x^3))}{(2x^2+O(x^3))}\\ &=\lim_{x\to 0} \dfrac{3x^2/2+O(x^3)}{2x^2+O(x^3)}=\frac34.}$$
Using L-Hospital Rule: $$\eqalign{\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}& =\lim_{x\to 0} \dfrac{3\cos^2 x.\sin x}{\sin2x+2x\cos2x}\\ &=\lim_{x\to 0} \dfrac{3\sin x}{\sin2x+2x\cos2x}\\ &=\lim_{x\to 0} \dfrac{3\cos x}{2\cos2x+2\cos2x-4x\sin2x}\\ &=\frac{3\times1}{2\times1+2\times1-4\times0}=\frac34.}$$
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$$\eqalign{ \lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x} &=\lim_{x \to 0} \dfrac{(1-\cos x)(1+\cos x+\cos^2 x)}{2x\sin x\cos x}\\ &=\lim_{x \to 0} \dfrac{3(1-\cos x)}{2x\sin x}\\ &=\lim_{x \to 0} \dfrac{6\sin^2(x/2)}{4x\sin(x/2)\cos(x/2)}\\ &=\lim_{x/2 \to 0} \dfrac{3\sin(x/2)}{4(x/2)}\\ &=\dfrac{3}{4}. }$$
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Hint:
You can break the limit into two parts:
$$\eqalign{ \lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x} & =\lim_{x\to0}\dfrac{\sin^2x+\cos^2x-\cos^3x}{x\sin2x} \\ &=\lim_{x\to0}\dfrac{\sin^2x+\cos^2x(1-\cos x)}{x\sin2x} \\ &=\lim_{x\to0}\color{darkmagenta}{\dfrac{\sin^2x}{x\sin2x}}+\color{darkblue}{\dfrac{\cos^2x(1-\cos x)}{x\sin2x}}. }$$
The first part can be written as $\color{darkmagenta}{\tfrac{\tan x}{2x}}$ and the second as $\color{darkblue}{\tfrac{{\cos x\tan\left(\tfrac x2\right)}}{2x}}$. Hopefully those two will be easier to deal with than the original limit. ;-)
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After two applications of L'Hôpital's rule: $$ \displaystyle \begin{aligned} \lim_{x \to 0} \left( \dfrac{1-\cos^3 (x)}{x\sin (2x)} \right) & \ = \ \lim_{x \to 0} \left( \dfrac{3\sin (x) \cos^2 (x)}{\sin (2x) + 2x\cos (2x)} \right) \\ & \ = \ \lim_{x \to 0} \left( \dfrac{3\cos^3 (x) - 6\cos (x) \sin (x)}{4\cos (2x) - 4x \sin (2x)} \right) = \ ? \end{aligned} $$
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By Taylor series $$\cos x \sim_0 1 - \frac{x^2}{2}, \,\sin 2x \sim_0 2x\, \text{ and }\,\Big(1-\frac{x^2}{2}\Big)^3 \sim_0 1 - \frac{3x^2}{2}$$ so $$\dfrac{1-\cos^3 x}{x\sin2x} \sim_0 \frac{1 - \Big(1-\frac{x^2}{2}\Big)^3}{2x^2} \sim_0 \frac{ \frac{3x^2}{2} }{2x^2} = \frac{3}{4}.$$
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$\begingroup$ Thanks for the explanation bro... well I got to learn Taylor series now.. :) $\endgroup$– AdarshCommented Aug 5, 2014 at 13:33
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$$\lim_{x \to 0}\frac{1-\cos^3x}{2x^2} \cdot \lim_{x \to 0}\frac{2x}{\sin(2x)}$$ The right limit is equal to 1. For the left limit, you may use the rule of de l'Hôpital: $$\eqalign{ \lim_{x \to 0}\frac{1-\cos^3x}{2x^2}&=\lim_{x \to 0}\frac{3\cos^2 x \sin x}{4x}\\ &=\lim_{x \to 0}\frac{-6\cos x \sin^2 x + 3 \cos^3 x}{4}\\ &=\frac{-6\cdot 0 + 3 \cdot 1}{4}\\ &=\frac{3}{4}. }$$
