Limits 2

hey try L'Hospital..

[(1+x)^1/x + e(x-1) ]/sin^-1x

This is very simple through that...

u could try taylor series * i havent tried it but i think that will do as well :)

naman were u able to reach any solution after my hint?

Why dont u try L'Hospital.. it will make it very simple.. only the derivation looks a bit tough.. it is not actually :)

hey try L'Hospital..

L= limit x-0 [(1+x)^1/x + e(x-1) ]/sin^-1x

t=(1+x)^1/x

L= limit x-0 [t + e(x-1) ]/sin^-1x
L= limit x-0 [t + e(x-1) ]/sin^-1x
L= (limit x->0) t/sin^-1x + (limit x-0) [e(x-1) ]/sin^-1x

(limit x->0) t/sin^-1x

dt/dx=t(1/x-1/(x+1) - ln(1+x)/x² ...... Try getting this!
= t(1/x-1/(x+1)- (x-x²+.. higher order terms)/x²
= t(-1/(x+1) + 1/2 + x(f(x))

now apply LH rule
(limit x->0) t/sin^-1x + [e(x-1) ]/sin^-1x

= (limx->0) t(-1/(x+1) + 1/2 + x(f(x)) √1-x² + e
= (limx->0) -e/2 + e
=e/2

typo if nay should by forgiven :)