The correct option is
A e2xlimy→∞(l+xy)2y=z
Now, when limit are applied, it is of indeterminate form 1∞, so, writing it after applying log on both sides,
limy→∞2ylog(1+xy)=logz
Now, LHS is of indeterminate form 0×∞ as limy→∞2y=∞ and
limy→∞log(1+xy)=0
So, Writing it as f(x)ϕ(x) where, f(x)=log(1+xy) and ϕ(x)=2y
limy→∞log(1+xy)12y=logyz
Now, it is of indeterminate 00, So applying L Hospital Rule,
limy→∞1(1+xy)(2xy2)(12y2)=logyz
limy→∞2x(1+xy)=logyz
Now applying limit 2x(1+0)=logyz
2x=logyz
zy=e2x
Hence e2x is the correct option