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Question

The value of limy(1+xy)2y is

A
e2x
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B
e2x
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C
ex
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D
ex
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Solution

The correct option is A e2x
limy(l+xy)2y=z

Now, when limit are applied, it is of indeterminate form 1, so, writing it after applying log on both sides,

limy2ylog(1+xy)=logz

Now, LHS is of indeterminate form 0× as limy2y= and

limylog(1+xy)=0

So, Writing it as f(x)ϕ(x) where, f(x)=log(1+xy) and ϕ(x)=2y

limylog(1+xy)12y=logyz

Now, it is of indeterminate 00, So applying L Hospital Rule,

limy1(1+xy)(2xy2)(12y2)=logyz

limy2x(1+xy)=logyz

Now applying limit 2x(1+0)=logyz

2x=logyz

zy=e2x

Hence e2x is the correct option

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