Limit of $\lim\limits_{x \to\infty} 3\left(\sqrt{\strut x}\sqrt{\strut x-3}-x+2\right)$ I have to compute this limit: $$\lim_{x \to\infty} 3(\sqrt{\strut x}\sqrt{\strut x-3}-x+2)$$ wolfram alpha says%2C%20x%20-%3E%20%5C%5BInfinity%5D%5D) that answer is $\frac{3}{2}$, but I can't get why. Does anyone know how to get this limit?

The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of $x$. \begin{align*} \lim_{x\to\infty}3\left(\sqrt{x}\sqrt{x-3} - x+2\right) &= 3\lim_{x\to\infty}\left(\sqrt{x^2-3x} - (x-2)\right)\\\ &= 3\lim_{x\to\infty}\frac{(\sqrt{x^2-3x}-(x-2))(\sqrt{x^2-3x}+(x-2))}{\sqrt{x^2-3x}+ (x-2)} \\\ &= 3\lim_{x\to\infty}\frac{(x^2-3x) - (x-2)^2}{\sqrt{x^2-3x}+(x-2)}\\\ &= 3\lim_{x\to\infty}\frac{x^2 - 3x - x^2 + 4x - 4}{\sqrt{x^2-3x} + (x-2)}\\\ &= 3\lim_{x\to\infty}\frac{x - 4}{\sqrt{x^2-3x}+(x-2)}\\\ &= 3\lim_{x\to\infty}\frac{\frac{1}{x}(x-4)}{\frac{1}{x}(\sqrt{x^2-3x}+(x-2))}\\\ &= 3\lim_{x\to\infty}\frac {1 - \frac{4}{x}}{\sqrt{1 - \frac{3}{x}} + 1 - \frac{2}{x})}\\\ &= 3\left(\frac{1}{\sqrt{1}+1}\right) = \frac{3}{2}. \end{align*}