lim_(xat 0) [cos^2 (4x)]/x^2 −lim_(x at 0) [cos^3 (6x)]/x^2


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Question Number 64539 by mmkkmm000m last updated on 19/Jul/19

${l i m}_{x a t 0} [{c o s}^{2} (4 x)] / x^{2} - {l i m}_{x a t 0} [{c o s}^{3} (6 x)] / x^{2}$
Commented by kaivan.ahmadi last updated on 19/Jul/19

${l i m}_{x \to 0} \frac{[{c o s}^{2} 4 x]}{x^{2}} - {l i m}_{x \to 0} \frac{[{c o s}^{3} 6 x]}{x^{2}} =$ ${l i m}_{x \to 0} \frac{[{c o s}^{2} 4 x] - [{c o s}^{3} 6 x]}{x^{2}} = \frac{[1^{-}] - [1^{-}]}{0^{+}} = \frac{0 - 0}{0^{+}} = \frac{0}{0^{+}} = 0$
Commented by MJS last updated on 19/Jul/19

$\frac{0}{0} \neq 0$
Commented by mathmax by abdo last updated on 19/Jul/19
$let A(x) =((cos^2 (4x)−cos^3 (6x))/x^2 ) lim_(x→0) A(x)? we have cosu ∼1−(u^2 /2) (u→0) and cos^2 x =((1+cos(2x))/2) ∼(1/2) +(1/2){1−((4x^2 )/2)} =1 −x^2 ⇒cos^2 (4x) ∼1−16x^2 cos^3 (6x) =cos(6x)cos^2 (6x) ∼(1−18x^2 )(1−36x^2 ) =1−36x^2 −18x^2 +18.36 x^4 ∼1−54x^2 ⇒ A(x) ∼ ((1−16x^2 −1+54x^2 )/x^2 ) =38 ⇒lim_(x→0) A(x)=38$
$l e t A (x) = \frac{{c o s}^{2} (4 x) - {c o s}^{3} (6 x)}{x^{2}} {l i m}_{x \to 0} A (x) ?$ $w e h a v e c o s u \sim 1 - \frac{u^{2}}{2} (u \to 0) a n d {c o s}^{2} x = \frac{1 + c o s (2 x)}{2}$ $\sim \frac{1}{2} + \frac{1}{2} {1 - \frac{4 x^{2}}{2}} = 1 - x^{2} \Rightarrow {c o s}^{2} (4 x) \sim 1 - 16 x^{2}$ ${c o s}^{3} (6 x) = c o s (6 x) {c o s}^{2} (6 x) \sim (1 - 18 x^{2}) (1 - 36 x^{2})$ $= 1 - 36 x^{2} - 18 x^{2} + 18.36 x^{4} \sim 1 - 54 x^{2} \Rightarrow$ $A (x) \sim \frac{1 - 16 x^{2} - 1 + 54 x^{2}}{x^{2}} = 38 \Rightarrow {l i m}_{x \to 0} A (x) = 38$
Commented by kaivan.ahmadi last updated on 19/Jul/19

$\frac{0}{ε} = 0$
Commented by kaivan.ahmadi last updated on 19/Jul/19

$s i r [] m e a n s i n t e g e r p a r t$
Commented by mathmax by abdo last updated on 19/Jul/19

$p e r h a p s s i r b u t y o u r a n s w e r i s n o t c o r r e c t . . .$
Commented by kaivan.ahmadi last updated on 19/Jul/19

$w h y ?$
Commented by kaivan.ahmadi last updated on 19/Jul/19

$\frac{0}{0} i s u n d e f i n e d$
Commented by MJS last updated on 19/Jul/19

$\lim_{x \to 0} [\cos^{2} 4 x] = 0$ $\lim_{x \to 0} [\cos^{3} 6 x] = 0$ $I think that in this case the difference of the$ $given limits is undefined$
	Answered by MJS last updated on 19/Jul/19

	$\lim_{x \to 0} \frac{\cos^{2} 4 x - \cos^{3} 6 x}{x^{2}} =$ $= \lim_{x \to 0} \frac{2 - \cos 18 x + 2 \cos 8 x - 3 \cos 6 x}{4 x^{2}}$ $= \lim_{x \to 0} \frac{\frac{d^{2}}{{d x}^{2}} [2 - \cos 18 x + 2 \cos 8 x - 3 \cos 6 x]}{\frac{d^{2}}{{d x}^{2}} [4 x^{2}]} =$ $= \frac{1}{2} \lim_{x \to 0} (81 \cos 18 x - 32 \cos 8 x + 27 \cos 6 x) =$ $= 38$
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