Lim tend to infinity (1+1/n)en

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lim x \to \infty \frac{x}{\sqrt{{4 x}^{2} + 1}} = \frac{1}{2}

The value of limit n tends to infinity ((1.5)ⁿ + {(1+0.0001)¹⁰⁰⁰⁰}ⁿ)^1/n

^{{} denotes Greatest Integer function is?}

^{Options: A.1 B.1/2 C.Does not exist D.2}

x (x + 2) = 3 - a x^{2}

a

y = x^{2}

lim x \to \infty \frac{\sqrt{{3 x}^{2} - 1} + \sqrt{{2 x}^{2} - 1}}{4 x + 3}

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