The value of limx→∞ 3×2−1+2×2−14x+3 ,is

# The value of $\underset{x\to \mathrm{\infty }}{lim} \frac{\sqrt{3{x}^{2}-1}+\sqrt{2{x}^{2}-1}}{4x+3}$ ,is

1. A

$\frac{\sqrt{3}-\sqrt{2}}{4}$

2. B

$\frac{1}{4\left(\sqrt{3}-\sqrt{2}\right)}$

3. C

$\frac{\sqrt{3}-\sqrt{2}}{2}$

4. D

none of these

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### Solution:

Dividing each term in the numerator and denominator by x, we get

$\begin{array}{l}\underset{x\to \mathrm{\infty }}{lim} \frac{\sqrt{3{x}^{2}-1}+\sqrt{2{x}^{2}-1}}{4x+3}\\ =\underset{x\to \mathrm{\infty }}{lim} \frac{\sqrt{3-1/{x}^{2}}+\sqrt{2-1/{x}^{2}}}{4+3/x}=\frac{\sqrt{3}+\sqrt{2}}{4}=\frac{1}{4\left(\sqrt{3}-\sqrt{2}\right)}\end{array}$

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