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

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4x^2+2x}-\sqrt[3]{8x^3+4x^2}$$

Commented by jagoll last updated on 28/Apr/20

$$thankyousir$$

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

$$\underset{x\to \mathrm{\infty }}{\lim }2x\sqrt{1+\frac{1}{2x}}-2x\sqrt[3]{1+\frac{1}{2x}}=$$$$[let\frac{1}{2x}=p\Rightarrow p\to 0]$$$$\underset{p\to 0}{\lim }\frac{\sqrt{1+p}-\sqrt[3]{1+p}}{p}=$$$$\underset{p\to 0}{\lim }\frac{(1+\frac{p}{2})-(1+\frac{p}{3})}{p}=$$$$\underset{p\to 0}{\lim }\frac{\left(\frac{p}{6}\right)}{p}=\frac{1}{6}$$

Commented by abdomathmax last updated on 28/Apr/20

$$letf\left(x\right)=\sqrt{4x^2+2x}{-}^3\sqrt{8x^3+4x^2}$$$$wehave\sqrt{4x^2+2x}=2x\sqrt{1+\frac{1}{2x}}\sim 2x(1+\frac{1}{4x})(x\to +\mathrm{\infty })$$$${(8x^3+4x^2)}^{\frac{1}{3}}=2x{(1+\frac{1}{2x})}^{\frac{1}{3}}\sim 2x(1+\frac{1}{6x})\Rightarrow$$$$f\left(x\right)\sim 2x+\frac{1}{2}-2x-\frac{1}{3}=\frac{1}{6}\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=\frac{1}{6}$$

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