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NIMCET 2018 Question Paper with solutions

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Question : 17 of 120

Marks: +1, -0

If

\lim\limits_{x\to\infty}

\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}

e^2

, then the value of a and b are

a ∊ R , b = 2
a = 1 , b ∊ R
a ∊ R , b ∊ R
a = 1 , b = 2

Solution:

Given that,

\lim\limits_{x\to\infty}

\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}

=

e^2

LHS is the form of

1{\infty} ,

therefore

\lim\limits_{x\to\infty} \left(1+\frac{1}{x}\right)^x

=

\lim\limits_{x\to\infty}

=

\lim\limits_{x\to\infty}

f (x)

Thus ,

\lim\limits_{x\to\infty}

\left(\frac{a}{x}+\frac{b}{x^2}\right)^{2x}

=

e^2

using property

⇒

\lim\limits_{x\to\infty} \left(2a+\frac{2b}{x}\right)

=

e^2

⇒

2^{2a+0}

=

e^2

On comparing exponent power both sides, we get

a = 1 , b ∊ R

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