<s>[INST] Calculate $\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$ $$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$ 
I know this limit must be equal to $\frac{1}{2}$ but I can't figure why. This is just one of the thing I tried to solve this limit:
$$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$
$$\lim_{x\to\infty}x\biggr(\sqrt{\frac{x}{x-1}}-1\biggr)$$ 
Now I try to evaluate the limit. I know that $\lim_{x\to\infty}\sqrt\frac{x}{x-1}$ is equal to 1 so that means the above limit evaluates to $\infty * 0$ which is indeterminate form. I do not know what to do next, would greatly appreciate some help.
 [/INST] Her is an elementary way:
\begin{eqnarray*} \biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)
& = & x \biggr(\sqrt{\frac{x}{x-1}}-1\biggr) \\
& = & x \biggr(\frac{\frac{x}{x-1}-1}{\sqrt{\frac{x}{x-1}}+1}\biggr) \\
& = & x \biggr(\frac{1}{(x-1)\sqrt{\frac{x}{x-1}}+(x-1)}\biggr) \\
& = & \frac{1}{(1-\frac{1}{x})\sqrt{1+\frac{1}{x-1}}+1-\frac{1}{x}} \\
& \stackrel{x \to \infty}{\longrightarrow} & \frac{1}{(1-0)\sqrt{1+0}+1-0}  = \frac{1}{2}
\end{eqnarray*}
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