If lim x -1-a-x-b-x22 x-e2- ther

The correct option is B a = 1, b Rx→∞lim ( 1+a/x+b/x^2 )^2x=e^2 ⇒x→∞lim ( 1+(ax+b)/x^2 )^2x=e^2 ⇒x→∞limax+b/x^2 must be equal to zero ⇒x→∞lim ( a/x+b/x^2 )=0 ...

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Byju's Answer

Standard XII

Mathematics

Range of Quadratic Expression

If lim x →∞1+...

Question

If $l i m x \to \infty {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2}$ , then

a = 1 , b = 2

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a = 2, b = 1

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a = 1, b

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None of these

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Solution

The correct option is C a = 1, b R
$l i m x \to \infty {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2} \Rightarrow l i m x \to \infty {(1 + \frac{(a x + b)}{x^{2}})}^{2 x} = e^{2} \Rightarrow l i m x \to \infty \frac{a x + b}{x^{2}} m u s t b e e q u a l t o z e r o \Rightarrow l i m x \to \infty (\frac{a}{x} + \frac{b}{x^{2}}) = 0 ∴ a a n d b ϵ R F r o m E q . (i), e^{l i m x \to \infty \frac{(a x + b)}{x^{2}} 2 x = e^{2}} \Rightarrow e^{l i m x \to \infty \frac{2 (a x + b)}{x} 2 x} = e^{2} \Rightarrow l i m x \to \infty (\frac{a}{x} + \frac{b}{x^{2}}) = 0 ∴ a = 1 a n d b ϵ R$

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If limx→∞(1+ax+bx2)2x=e2, then

The correct option is C a = 1, b Rlimx→∞(1+ax+bx2)2x=e2⇒limx→∞(1+(ax+b)x2)2x=e2⇒limx→∞ax+bx2 must be equal to zero⇒limx→∞(ax+bx2)=0∴a and bϵRFrom Eq.(i),elimx→∞(ax+b)x22x=e2⇒elimx→∞2(ax+b)x2x=e2⇒limx→∞(ax+bx2)=0∴a=1 and bϵR

If $l i m x \to \infty {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2}$ , then