How to solve this limit, normal methods result in zero. Online sources say otherwise. in one part of my homework, we are asked to solve this limit:
$$
\lim_{x\to\infty}\sqrt{\frac{x^3}{x-3}} - x
$$
The result should be $3/2$, but when I try it, I always get $0$ instead.
Please, do not post the full solution, I would only like to be pointed in the right direction.
ps:
Right now I got to:
$$
\lim_{x\to\infty}\frac{x\sqrt{x^2-3x} - x^2 + 3x}{x-3}
$$
EDIT:
Thanks everyone, now I understand it :) . Marking as solved!

We have
\begin{align} \sqrt{\frac{x^3}{x - 3}} - x &=  x\sqrt{\frac{x}{x - 3}} - x\\
&=  x\left(\sqrt{\frac{1}{1 - \frac{3}{x}}} - 1\right)\\
&= x\left(\frac{1}{\sqrt{1 - \frac{3}{x}}} - 1\right)\\
&=  \frac{x}{\sqrt{1 - \frac{3}{x}}}\left(1 - \sqrt{1 - \frac{3}{x}}\right)\\
&= \frac{x}{\sqrt{1 - \frac{3}{x}}}\frac{\frac{3}{x}}{1 + \sqrt{1 - \frac{3}{x}}}\\
&= \frac{3}{\sqrt{1 - \frac{3}{x}}} \frac{1}{1 + \sqrt{1 -\frac{3}{x}}}
\end{align}
Since $3/x \to 0$ as $x\to \infty$, the last expression tends to $$\frac{3}{1}\cdot\frac{1}{2} = \frac{3}{2}$$
Hence, your limit is $3/2$.
