⊚BeMath⊚ lim_(x→0) ((2−3cos^6 x cos^4 2x cos^2 4x+cos x)/(36x^2 ))


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Question Number 107414 by bemath last updated on 10/Aug/20

$⊚ B e M a t h ⊚$ $\lim_{x \to 0} \frac{2 - 3 \cos^{6} x \cos^{4} 2 x \cos^{2} 4 x + \cos x}{36 x^{2}}$
	Answered by ajfour last updated on 10/Aug/20
	$L=lim_(x→0) ((2−3(1−sin^2 x)^3 (1−sin^2 2x)^2 (1−sin^2 4x)+cos x)/(36x^2 )) = lim_(x→0) ((2−3(1−3sin^2 x)(1−2sin^2 2x)(1−sin^2 4x)+cos x)/(36x^2 )) = lim_(sin x=h→0) ((2−3(1−3h^2 ){1−8h^2 (1−h^2 )}{1−4[4(h^2 )(1−h^2 )](1−2h^2 )^2 }+(√(1−h^2 )))/(36h^2 )) =lim_(h^2 =k→0) ((2−3{1−3k){1−8k(1−k){1−16k+...}+(√(1−k)))/(36k)) = lim_(k→0) ((2−3(1−3k)(1−8k+..)(1−16k+..)+(√(1−k)))/(36k)) = lim_(k→0) ((2−3(1−27k+...)+1−(k/2))/(36k)) = ((81−(1/2))/(36)) = ((161)/(72)) .$
	$L = \lim_{x \to 0} \frac{2 - 3 {(1 - \sin^{2} x)}^{3} {(1 - \sin^{2} 2 x)}^{2} (1 - \sin^{2} 4 x) + \cos x}{36 x^{2}}$ $= \lim_{x \to 0} \frac{2 - 3 (1 - 3 \sin^{2} x) (1 - 2 \sin^{2} 2 x) (1 - \sin^{2} 4 x) + \cos x}{36 x^{2}}$ $= \lim_{\sin x = h \to 0} \frac{2 - 3 (1 - 3 h^{2}) {1 - 8 h^{2} (1 - h^{2})} {1 - 4 [4 (h^{2}) (1 - h^{2})] {(1 - 2 h^{2})}^{2}} + \sqrt{1 - h^{2}}}{36 h^{2}}$ $= \lim_{h^{2} = k \to 0} \frac{2 - 3 {1 - 3 k) {1 - 8 k (1 - k) {1 - 16 k + . . .} + \sqrt{1 - k}}{36 k}$ $= \lim_{k \to 0} \frac{2 - 3 (1 - 3 k) (1 - 8 k + . .) (1 - 16 k + . .) + \sqrt{1 - k}}{36 k}$ $= \lim_{k \to 0} \frac{2 - 3 (1 - 27 k + . . .) + 1 - \frac{k}{2}}{36 k}$ $= \frac{81 - \frac{1}{2}}{36} = \frac{161}{72} .$
	Answered by john santu last updated on 11/Aug/20

	$⋇ JS ⋇$ $\lim_{x \to 0} \frac{2 - 3 {(1 - \frac{x^{2}}{2})}^{6} {(1 - \frac{4 x^{2}}{2})}^{4} {(1 - \frac{16 x^{2}}{2})}^{2} + 1 - \frac{x^{2}}{2}}{36 x^{2}} =$ $\lim_{x \to 0} \frac{3 - 3 (1 - 3 x^{2}) (1 - 8 x^{2}) (1 - 16 x^{2}) - \frac{x^{2}}{2}}{36 x^{2}} =$ $s e t x^{2} = t$ $\lim_{t \to 0} \frac{3 - 3 (1 - 3 t) (1 - 8 t) (1 - 16 t) - \frac{t}{2}}{36 t} =$ $\frac{1}{36} \lim_{t \to 0} \frac{3 - 3 (1 - 11 t + 24 t^{2}) (1 - 16 t) - \frac{t}{2}}{t} =$ $\frac{1}{36} \lim_{t \to 0} \frac{3 - 3 (1 - 27 t + o (t)) - \frac{t}{2}}{t} =$ $\frac{1}{36} \lim_{t \to 0} \frac{81 t - \frac{t}{2} + 3 o (t)}{t} = \frac{1}{36} \times (\frac{161}{2})$ $= \frac{161}{72}$
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