lmmits

given limit = e^{limx→a}[2-x/a -1 ]tan πa/2x [ the limit is in the power of e]

limx→a {1-x/a}tan πa/2x =limx→a tan πx/2a / a/a-x this is of the form ∞/∞ and can be evaluated by l hospital`s rule.

i think the limit does not exist as on continuosly using l`hospital`s rule its remaining in ∞/∞ form

whats the answer rohit?

even i got that.........though i wasnt sure..as it was a skul paper.............dunno d answer,,,,dude,,,thanxn....

i m getting e^2/Î

hint take tanx=sinx/cosx.....then apply l hospitals to (1-x/a)/cosx (0/0 form)....it will be 2/Î ...hence the answer..

Lim_x→a [1 + (1- x/a)]^{tan (πx/2a)}

= Lim_{1-x/a→0} [1 + (1- x/a)]^{1/(1-x/a)tan (πx/2a) (1-x/a)}

= e^{Lim _x→a tan (πx/2a) (1-x/a)}

Now, Lim _x→a tan (πx/2a) (1-x/a)

= Lim_{π/2 (1-x/a)→0} [π/2 (1-x/a)]/[tan π/2 (1-x/a)] * 2/π

= 2/Ï€

Hence, the given limit is 2/Ï€

kya bhaiya! mere answer ko kabhi kabhi pink kar dijiye...
...:)
arey bhai! confidence badhta hai...:)