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

=limx→∞ 1+ax+bx22x =elimx→∞ 2x1+ax+bx2-1 =elimx→∞ 2xax+bx2=limx→∞ (e)2(a+b/x) =e2ae2a=e2⇒a=1 and b∈R

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

=limx→∞ 1+ax+bx22x =elimx→∞ 2x1+ax+bx2-1 =elimx→∞ 2xax+bx2=limx→∞ (e)2(a+b/x) =e2ae2a=e2⇒a=1 and b∈R

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

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$a \in R, b \in R$

$a = 1, b \in R$

$a \in R, b = 2$

$a = 1, b = 2$

answer is B.

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Detailed Solution

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

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

$e^{2 a} = e^{2} \Rightarrow a = 1 and b \in R$

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

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