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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)

best_answer

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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