lim-x-4x-2-2x-8x-3-4x-2-1-3-

Question Number 91158 by jagoll last updated on 28/Apr/20

\lim_{x \to \infty} \sqrt{4 x^{2} + 2 x} - \sqrt[3]{8 x^{3} + 4 x^{2}}

Commented by jagoll last updated on 28/Apr/20

t h a n k y o u s i r

Commented by john santu last updated on 28/Apr/20

\lim_{x \to \infty} 2 x \sqrt{1 + \frac{1}{2 x}} - 2 x \sqrt[3]{1 + \frac{1}{2 x}} =

[l e t \frac{1}{2 x} = p \Rightarrow p \to 0]

\lim_{p \to 0} \frac{\sqrt{1 + p} - \sqrt[3]{1 + p}}{p} =

\lim_{p \to 0} \frac{(1 + \frac{p}{2}) - (1 + \frac{p}{3})}{p} =

\lim_{p \to 0} \frac{(\frac{p}{6})}{p} = \frac{1}{6}

Commented by abdomathmax last updated on 28/Apr/20

l e t f (x) = \sqrt{4 x^{2} + 2 x} -^{3} \sqrt{8 x^{3} + 4 x^{2}}

w e h a v e \sqrt{4 x^{2} + 2 x} = 2 x \sqrt{1 + \frac{1}{2 x}} \sim 2 x (1 + \frac{1}{4 x}) (x \to + \infty)

{(8 x^{3} + 4 x^{2})}^{\frac{1}{3}} = 2 x {(1 + \frac{1}{2 x})}^{\frac{1}{3}} \sim 2 x (1 + \frac{1}{6 x}) \Rightarrow

f (x) \sim 2 x + \frac{1}{2} - 2 x - \frac{1}{3} = \frac{1}{6} \Rightarrow {l i m}_{x \to + \infty} f (x) = \frac{1}{6}