# I have to compute this limit: $$\displaystyle\lim_{{{x}\to\infty}}{3}{\left(\sqrt{{{x}}}\sqrt{{{x}-{3}}}-{x}+{2}\right)}$$

I have to compute this limit:
$\underset{x\to \mathrm{\infty }}{lim}3\left(\sqrt{x}\sqrt{x-3}-x+2\right)$
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Luciana Cline
The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of x
$\underset{x\to \mathrm{\infty }}{lim}3\left(\sqrt{x}\sqrt{x-3}-x+2\right)=3\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{{x}^{2}-3x}-\left(x-2\right)\right)$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{\left(\sqrt{{x}^{2}-3x}-\left(x-2\right)\right)\left(\sqrt{{x}^{2}-3x}+\left(x-2\right)\right)}{\sqrt{{x}^{2}-3x}+\left(x-2\right)}$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{\left({x}^{2}-3x\right)-{\left(x-2\right)}^{2}}{\sqrt{{x}^{2}-3x}+\left(x-2\right)}$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{2}-3x-{x}^{2}+4x-4}{\sqrt{{x}^{2}-3x}+\left(x-2\right)}$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{x-4}{\sqrt{{x}^{2}-3x}+\left(x-2\right)}$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{x}\left(x-4\right)}{\frac{1}{x}\left(\sqrt{{x}^{2}-3x}+\left(x-2\right)\right)}$
$=3\underset{x\to \mathrm{\infty }}{lim}\frac{1-\frac{4}{x}}{\sqrt{1-\frac{3}{x}}+1-\frac{2}{x}}\right\}$
$=3\left(\frac{1}{\sqrt{1}+1}\right)=\frac{3}{2}$
Jeffrey Jordon