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Mathematics 16 Online
OpenStudy (anonymous):

Evaluate the limit, if it exists: lim (as x->1) tan^-1 x ---------- sin^-1x +1

OpenStudy (anonymous):

\[\frac{ \arctan(1) }{ \arcsin(1)+1 }=\frac{ \pi/4 }{ \pi/2+1 }\] you can simplify further more

OpenStudy (anonymous):

unless, of course, you mean the function is:\[\frac{ \tan^-1(x) }{ \sin^-1(x+1) }\] this way the limit does not exist cause 2 is in the domain of arcsin

OpenStudy (anonymous):

Wow I feel stupid lol. I knew it was an inverse but I was trying to do tan (1) and sin (1) instead xD. Thank you for the help!

OpenStudy (anonymous):

And no it's just x, not x+1 :P

OpenStudy (anonymous):

okay that's good then :D , also you are very welcome ;)

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