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Question Number 200741 by Rydel last updated on 22/Nov/23

lim_(x→a) ((asin x−xsin a)/(x−a))

limxaasinxxsinaxa

Commented by JDamian last updated on 22/Nov/23

really?

Answered by JDamian last updated on 22/Nov/23

hint: xsin a−asin a=(x−a)sin a

hint:xsinaasina=(xa)sina

Commented by Rydel last updated on 22/Nov/23

lim_(x→a) ((asin x−xsina )/(x−a))

limxaasinxxsinaxa

Answered by MM42 last updated on 22/Nov/23

hop→lim_(x→a)  ((acosx−sina)/1)  =acosa−sina ✓

hoplimxaacosxsina1=acosasina

Answered by tri26112004 last updated on 23/Nov/23

= lim_(x→a)  ((a(sin x − sin a)+asin a−xsin a)/(x−a))  = lim_(x→a)  ((2acos(((x+a)/2))sin(((x−a)/2)))/(x−a))+lim_(x→a)  (((a−x)sin a)/(x−a))  = lim_(x→a)  acos(((x+a)/2)) − lim_(x→a)  sin a  = acos a − sin a

=limxaa(sinxsina)+asinaxsinaxa=limxa2acos(x+a2)sin(xa2)xa+limxa(ax)sinaxa=limxaacos(x+a2)limxasina=acosasina

Commented by Rydel last updated on 23/Nov/23

thank you very much

thankyouverymuch

Answered by MM42 last updated on 23/Nov/23

x−a=u  ⇒lim_(u→0)  ((asin(u+a)−(u+a)sina)/u)  =lim_(u→0)  ((asinucosa+asinacosu−usina−asina)/u)  =lim_(u→0) ( ((sinu)/u)acosa−sina−((1−cosu)/u)asina)  =acosa−sina

xa=ulimu0asin(u+a)(u+a)sinau=limu0asinucosa+asinacosuusinaasinau=limu0(sinuuacosasina1cosuuasina)=acosasina

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