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Question Number 82541 by zainal tanjung last updated on 22/Feb/20

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

Commented by john santu last updated on 22/Feb/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4x^2-8x+3}-(2x+4)=$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4x^2-8x+3}-\sqrt{4x^2+16x+16}=$$$$\frac{-8-16}{2.2}=-6$$

Commented by mathmax by abdo last updated on 22/Feb/20

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

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