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Question Number 105104 by bemath last updated on 26/Jul/20

$$\underset{x\to 0}{\lim }\frac{1-\cos {}^5\left(x\right)\cos {}^3\left(2x\right)\cos {}^3\left(3x\right)}{22x^2}?$$

Answered by bramlex last updated on 26/Jul/20

$$\underset{x\to 0}{\lim }\frac{5\cos {}^4\left(x\right)\sin \left(x\right)\cos {}^3\left(2x\right)\cos {}^3\left(3x\right)+6\cos {}^2\left(2x\right)\cos {}^5\left(x\right)\cos {}^3\left(3x\right)\sin \left(2x\right)+9\cos {}^2\left(3x\right)\cos {}^5\left(x\right)\cos {}^3\left(2x\right)\sin \left(3x\right)}{44x}=$$$$\frac{5+12+27}{44}=1.▴$$

Answered by bobhans last updated on 26/Jul/20

$$\cos x=\sqrt{1-\sin {}^{2}x}\approx 1-\frac{x^2}{2}$$$$\underset{x\to 0}{\lim }\frac{1-{(1-\frac{x^2}{2})}^5{(1-2x^2)}^3{(1-\frac{9x^2}{2})}^3}{22x^2}=$$$$\underset{x\to 0}{\lim }\frac{1-(1-\frac{5x^2}{2})(1-6x^2)(1-\frac{27x^2}{2})}{22x^2}=$$$$\underset{x\to 0}{\lim }\frac{1-(1-(\frac{5}{2}+6+\frac{27}{2})x^2+o\left(x^2\right))}{22x^2}=$$$$\underset{x\to 0}{\lim }\frac{(\frac{5}{2}+6+\frac{27}{2})x^2+o\left(x^2\right)}{22x^2}=\frac{44}{2\left(22\right)}=1$$$$(B⊚B)$$

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