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Question Number 82485 by naka3546 last updated on 21/Feb/20

Commented by abdomathmax last updated on 21/Feb/20
$let f(x)=(6/t^2 )−((sin(6t))/(t^3 cos^2 (3t))) ⇒ f(x)=(1/t^2 ){6−((sin(6t))/(t cos^2 (3t)))} we have sin(6t)∼6t cos^2 (3t)=((1+cos(6t))/2) ∼(1/2) +(1/2)(1−((36t^2 )/2)) =(1/2)+(1/2)−9t^2 =1−9t^2 ⇒ f(x)∼(1/t^2 ){6−((6t)/(t(1−9t^2 )))} =(1/t^2 ){6−(6/(1−9t^2 ))} =(6/t^2 )×((1−9t^2 −1)/((1−9t^2 ))) =((−54)/(1−9t^2 )) →−54 ⇒ lim_(x→0) f(x)=−54$
$l e t f (x) = \frac{6}{t^{2}} - \frac{s i n (6 t)}{t^{3} {c o s}^{2} (3 t)} \Rightarrow$ $f (x) = \frac{1}{t^{2}} {6 - \frac{s i n (6 t)}{t {c o s}^{2} (3 t)}} w e h a v e$ $s i n (6 t) \sim 6 t$ ${c o s}^{2} (3 t) = \frac{1 + c o s (6 t)}{2} \sim \frac{1}{2} + \frac{1}{2} (1 - \frac{36 t^{2}}{2})$ $= \frac{1}{2} + \frac{1}{2} - 9 t^{2} = 1 - 9 t^{2} \Rightarrow$ $f (x) \sim \frac{1}{t^{2}} {6 - \frac{6 t}{t (1 - 9 t^{2})}} = \frac{1}{t^{2}} {6 - \frac{6}{1 - 9 t^{2}}}$ $= \frac{6}{t^{2}} \times \frac{1 - 9 t^{2} - 1}{(1 - 9 t^{2})} = \frac{- 54}{1 - 9 t^{2}} \to - 54 \Rightarrow$ ${l i m}_{x \to 0} f (x) = - 54$
Commented by jagoll last updated on 22/Feb/20

$\lim_{t \to 0} \frac{6 t \cos^{2} t - \sin 6 t}{t^{3} \cos^{2} t} =$ $\lim_{t \to 0} \frac{1}{\cos^{2} t} \times \lim_{t \to 0} \frac{6 \cos^{2} t - 6 t \sin 2 t - 6 \cos 6 t}{3 t^{2}}$ $= 1 \times \lim_{t \to 0} \frac{- 6 \sin 2 t - (6 \sin 2 t + 12 t \cos 2 t) + 36 \sin 6 t}{6 t}$ $= \lim_{t \to 0} \frac{- 24 \cos 2 t - (12 \cos 2 t - 24 t \sin 2 t) + 216 \cos 6 t}{6}$ $= \frac{- 36 + 216}{6} = - 6 + 36 = 30 ?$
Commented by naka3546 last updated on 22/Feb/20

$\cos t o r \cos 3 t ?$
Commented by john santu last updated on 22/Feb/20

$t h i s a n s w e r i s - 18$
	Answered by john santu last updated on 22/Feb/20

	$\lim_{t \to 0} \frac{6}{t^{2}} - \frac{2 \sin 3 t \cos 3 t}{t^{3} \cos^{2} 3 t} =$ $\lim_{t \to 0} \frac{6}{t^{2}} - \frac{2 \tan 3 t}{t^{3}} = \lim_{t \to 0} \frac{6 t - 2 \tan 3 t}{t^{3}}$ $= \lim_{t \to 0} \frac{6 - 6 \sec^{2} 3 t}{3 t^{2}} = \lim_{t \to 0} \frac{2 (1 - \frac{1}{\cos^{2} 3 t})}{t^{2}}$ $= 2 \times \lim_{t \to 0} \frac{\cos^{2} 3 t - 1}{t^{2} \cos^{2} 3 t} = 2 \times \lim_{t \to 0} \frac{1}{\cos^{2} 3 t} \times \lim_{t \to 0} \frac{- \sin^{2} 3 t}{t^{2}}$ $= 2 \times 1 \times (- 9) = - 18$
	Answered by Henri Boucatchou last updated on 22/Feb/20

	$A t V (0), s i n 6 t \sim 6 t, \frac{6}{t^{2}} - \frac{s i n 6 t}{t^{3} (1 - {s i n}^{2} 3 t)} \sim \frac{6}{t^{2}} (1 - \frac{1}{1 - {(3 t)}^{2}})$ $\sim \frac{6}{t^{2}} (\frac{1 - 9 t^{2} - 1}{1 - 9 t^{2}}) \sim - 54 \frac{1}{1 - 9 t^{2}}$ $\Rightarrow \underset{x \to 0}{l i m} (\frac{6}{t^{2}} - \frac{s i n 6 t}{t^{3} {c o s}^{2} t}) = \underset{x \to 0}{l i m} (- 54 \frac{1}{1 - 9 t^{2}}) = - 54$
	Commented by john santu last updated on 22/Feb/20

	$n o t c o r r e c t \lim_{x \to 0} 1 - \sin^{2} 3 t \sim 1 - {(3 t)}^{2}$
	Commented by Henri Boucatchou last updated on 22/Feb/20

	$S o r r y, t a k e t = x$
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