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?

yah..i proved by both ways..that limits doesnt exist..make it of d form

1 + f(x).........

i m getting e^2/Î

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€

okkies...samaj gaya..thanx bro.......

thanks DUDE!

thanks buddy!!!!!!!!!!!!!!