If \lim_{x\rightarrow \infty }\left ( 1+\frac{a}{x}+\frac{b}{x^{2}} \right )^{2x}= e^{2}, then the values of a \: and \: b, are

  • Option 1)

    a,\in \: R,b=2

  • Option 2)

    a=\: 1,b\in R

  • Option 3)

    a,\in \: R,b\in R

  • Option 4)

    a=1\: and\: b=2

 

Answers (1)

As we learnt in 

1 to the power of infinity Form -

Let\:\:\;\lim_{x\rightarrow a}f(x)^{g(x)}\:\;\:where

f(a)=1\:\:\:and \;\:\:g(a)=\infty

Then\:\:\:\:e^\lim_{x\rightarrow a}(f(x)-1)g(x)

-

 

 \lim_{x\rightarrow \infty} \left ( 1+\frac{a}{x}+\frac{b}{x^{2}} \right )^{2x}=e^{2}

it is 1^{\infty } form

\lim_{x\rightarrow \infty} \left ( 1+\frac{a}{x} +\frac{b}{x^{2}}-1\right )\times 2x

\lim_{x\rightarrow \infty} \left(a+\frac{b}{x} \right )\times 2

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

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

\Rightarrow \left ( a+\frac{b}{x} \right )=1

so that a=1 and b\epsilon R


Option 1)

a,\in \: R,b=2

Incorrect option

Option 2)

a=\: 1,b\in R

Correct option

Option 3)

a,\in \: R,b\in R

Incorrect option

Option 4)

a=1\: and\: b=2

Incorrect option

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