If limx→∞ 1+ax+bx22x=e2, then

A
$a = 1, b = 2$
B
$a = 2, b = 1$
C
$a = 1, b \in R$
D
$a = b = 1$

Solution:

We have,

$\begin{array}{l} lim_{x \to \infty} {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} \\ = e^{lim_{x \to \infty} (\frac{a}{x} + \frac{b}{x^{2}}) \times 2 x} = e^{lim_{x \to \infty} (2 a + \frac{2 b}{x})} = e^{2 a} \\ ∴ lim_{x \to \infty} {(1 + \frac{a}{x} + \frac{b}{x^{2}})}^{2 x} = e^{2} \Rightarrow e^{2 a} = e^{2} \Rightarrow 2 a = 2 \Rightarrow a = 1 . \end{array}$

Hence, $a = 1$ and $b \in R$

$lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{''} (x)}{g^{''} (x)}$

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

Solution:

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