$$\lim_{x \rightarrow \infty} x[\sin ^{-1}(\frac{ x+1 }{ 2x+1 }) - \frac{ \Pi }{ 6 } ]$$

Question

amistre64 · Answer

... Lhop rule?  is my first thought

anonymous · Answer

First we have to have to indeterminate form..ryt

amistre64 · Answer

yes; 0/0 format is good

anonymous · Answer

But I did Everything but i never got the answer...

amistre64 · Answer

the 1/x tho might prove futile tho.  since the derivatices all have and x under

amistre64 · Answer

never tried it out on inf/inf tho

amistre64 · Answer

$$[\sin ^{-1}(\frac{ x+1 }{ 2x+1 }) - \frac{ \Pi }{ 6 } ]'=\frac{1}{\sqrt{1-\left(\frac{ x+1 }{ 2x+1 }\right)^2}}\frac{-1}{4x^2+4x+1}$$
$$[x^{-1}]'=-x^{-2}$$
does that stuff simplify any?

anonymous · Answer

$${d\over dx}x\left[\sin^{-1}\left(x+1\over 2x+1\right)-{\pi\over6}\right]\
=\left[\sin^{-1}\left(x+1\over 2x+1\right)-{\pi\over6}\right]+x\frac{1}{\Large\sqrt{1-\left({x+1\over2x+1}\right)^2}}
$$

anonymous · Answer

Aha..thxx guys i found my mistake.......