If $lim x \to 0 \frac{a x e^{x} - b log (1 + x)}{x^{2}} = 3$ then the values of $a, b$ are respectively

2, 2

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1, 2

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2, 1

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2, 0

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Solution

The correct option is A $2, 2$
$L = lim x \to 0 \frac{a x e^{x} - b log (1 + x)}{x^{2}}$
Applying L'Hospital rule,
$L = lim x \to 0 \frac{a e^{x} + a x e^{x} - \frac{b}{1 + x}}{2 x}$
For the limit to exist at $x = 0$ , we should have
$a + 0 - b = 0 \Rightarrow a = b$

Applying L'Hospital rule again, we get
$L = lim x \to 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{a e^{x} (x + 1) + a e^{x} + \frac{b}{(1 + x)^{2}}}{2} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$= \frac{2 a + b}{2}$
$= \frac{2 a + a}{2}$
$= \frac{3 a}{2}$
From the question,
$3 = \frac{3 a}{2}$
$\Rightarrow a = 2$

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