lim_(x→0) ((sin x −sin (sin x))/x^3 ) = ?


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 34458 by rahul 19 last updated on 06/May/18

$\lim_{x \to 0} \frac{\sin x - \sin (\sin x)}{x^{3}} = ?$
Commented by math khazana by abdo last updated on 06/May/18

$w e h a v e s i n x = x - \frac{x^{3}}{3!} + 0 (x^{5}) (x \to 0) a n d$ $s i n (s i n x) = s i n (x - \frac{x^{3}}{3!} + 0 (x^{5}))$ $= x - \frac{x^{3}}{3!} - {(x - \frac{x^{3}}{3!})}^{3} \times \frac{1}{3!} + 0 (x^{9}) \Rightarrow$ $s i n x - s i n (s i n x) = \frac{1}{3!} (x^{3} - 3 x^{2} \frac{x^{3}}{3!} + . . . .) + 0 (x^{9}) \Rightarrow$ ${l i m}_{x \to 0} \frac{s i n x - s i n (s i n x)}{x^{3}} = \frac{1}{3!} = \frac{1}{6} .$
Commented by math khazana by abdo last updated on 06/May/18

$l e t u s e h o s p i t a l t h e o r e m u (x) = s i n x - s i n (s i n x)$ $a n d v (x) = x^{3} w e h a v e u^{^{'}} (x) = c o s x - c o s x . c o s (s i n x)$ $u^{^{″}} (x) = - s i n x - (- s i n x c o s (s i n x) - {c o s x}^{2} s i n (s i n x))$ $= - s i n x + s i n x c o s (s i n x) + {c o s}^{2} x s i n (s i n x)$ $u^{(3)} (x) = - c o s x + c o s x c o s (s i n x) - {s i n}^{2} x s i n (s i n x)$ $- 2 c o s x s i n x s i n (s i n x) + {c o s}^{2} x c o s x c o s (s i n x)$ $u^{(3)} (0) = 1 a l s o v^{^{'}} (x) = 3 x^{2}, v^{^{″}} (x) = 6 x v^{(3)} (x) = 6$ $s o {l i m}_{x \to 0} \frac{s i n x - s i n (s i n x)}{x^{3}} = \frac{1}{6}$
Commented by math khazana by abdo last updated on 06/May/18

$d u e t o h o s p i t a l t h e o r e m m y a n s w e r i s c o r r e c t .$
Commented by rahul 19 last updated on 07/May/18
Thank you sir !
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18
	$=((lim )/(x→0)){2cos(((x+sinx)/2)).sin(((x−sinx)/2))}/x^3 =((lim)/(x→0))2.cos0.sin(((x−sinx)/2))/x^3 sinx=(x/1)−(x^3 /(3!))+(x^5 /(5!))−... =((lim)/(x→0))2.sin((x^3 /(3!)))/x^3 =((lim)/(x→0))((2sin((x^3 /(3!))))/((x^3 /(3!))×3!)) =(2/(3!^ )) =2/6=1/3$
	$= \frac{l i m}{x \to 0} {2 c o s (\frac{x + s i n x}{2}) . s i n (\frac{x - s i n x}{2})} / x^{3}$ $= \frac{l i m}{x \to 0} 2 . c o s 0 . s i n (\frac{x - s i n x}{2}) / x^{3}$ $s i n x = \frac{x}{1} - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . .$ $= \frac{l i m}{x \to 0} 2 . s i n (\frac{x^{3}}{3!}) / x^{3}$ $= \frac{l i m}{x \to 0} \frac{2 s i n (\frac{x^{3}}{3!})}{\frac{x^{3}}{3!} \times 3!}$ $= \frac{2}{3 \overset{}{!}}$ $= 2 / 6 = 1 / 3$
	Commented by MJS last updated on 06/May/18

	$something went wrong . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/May/18

	$y e s i g o t i t . . .$ $s i n (\frac{x - s i n x}{2})$ $i t s v a l u e i s s i n (\frac{x^{3}}{3! \times 2}) b u t b y m i s t a k e i p u t i t a s s i n$ $s i n (\frac{x^{3}}{3!})$ $s o c o r r e c t a n s i s$ $= \frac{l i m}{x \to 0} \frac{2 s i n (\frac{x^{3}}{3! \times 2})}{\frac{x^{3}}{3! \times 2} \times 3! \times 2}$ $= \frac{2}{3! \times 2}$ $= \frac{1}{6}$
	Commented by rahul 19 last updated on 07/May/18
	Thank you sir! There is also an option for "edit post" ��
	Answered by MJS last updated on 06/May/18

	$f (x) = \sin x - \sin \sin x$ $g (x) = x^{3}$ $f^{‴} (x) = (\cos^{3} x + \cos x) \cos \sin x + 3 \sin x \cos x \sin \sin x - \cos x$ $g^{‴} (x) = 6$ $f^{‴} (0) = 1$ $g^{‴} (0) = 6$ $Missing \left or extra \right$
	Commented by rahul 19 last updated on 07/May/18
	Thank you sir !
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum