Question

Finding a limit

Answer 1

No L'Hopital
\[\lim_{x \rightarrow 0} \frac{ arcsin(x) }{2tan(x) }\]

Answer 2

\[\lim_{x \rightarrow 0} \frac{ x\cos(x)*\arcsin(x) }{ 2x\sin(x) }\]
\[x=\sin(t)\]
\[t=\arcsin(x)\]
\[t \rightarrow 0 , x \rightarrow 0\]
Can I do something like this?
\[\lim_{x \rightarrow 0} \frac{ x\cos(x)*\arcsin(x) }{ 2x\sin(x) }=\frac{ 1 }{ 2} \lim_{x \rightarrow 0}\frac{ \cos(x) *x }{ \sin(x) }*\lim_{t \rightarrow 0}\frac{ t }{ \sin(t) }=\frac{ 1 }{ 2 }*1*1*1=\frac{ 1 }{ 2 }\]

Answer 3

how did u get an extra x ?

Answer 4

in the numerator...

Answer 5

if its already present, then its very easy, no need to substitute anything...

Answer 6

I multiplied numerator and denominator by x

Answer 7

ohh...

Answer 8

so your limit reduces to arcsin x / 2x
because denominator 2tanx/ x will be 2

Answer 9

now you can put x = sin theta if u want.

Answer 10

getting that ^ ?

Answer 11

oh, u did the same thing....and its correct...

Answer 12

Well, thanks for checking my work then, I guess lol

Answer 13

hmm....welcome ^_^