If Ltx→∞1+ax+bx22x=e2 then the values of a and b

Ltx→∞1+ax+bx22x=e2⇒elimx→∞(1+ax+bx2-1)2x =e2⇒elimx→∞(2a+2bx)=e2⇒ e2a=e2∴a=1 and b∈R.

If Ltx→∞1+ax+bx22x=e2 then the values of a and b

Ltx→∞1+ax+bx22x=e2⇒elimx→∞(1+ax+bx2-1)2x =e2⇒elimx→∞(2a+2bx)=e2⇒ e2a=e2∴a=1 and b∈R.

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If $\underset{x \to \infty}{L t} {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2}$ then the values of $a$ and $b$

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$a \in R, b \in R$

$a = 1 a n d b = 2$

$a \in R, b = 2$

$a = 1, b \in R$

answer is D.

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Detailed Solution

$\underset{x \to \infty}{L t} {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2}$

$\Rightarrow$ $e^{\lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x^{2}} - 1) 2 x^{}} = e^{2}$

$\Rightarrow e^{\lim_{x \to \infty} (2 a + \frac{2 b}{x})} = e^{2}$

$\Rightarrow e^{2 a} = e^{2}$

$∴ a = 1 a n d b \in R .$

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If $\underset{x \to \infty}{L t} {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2}$ then the values of $a$ and $b$