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.
 A: With $t:= \frac{x}{x-1}$ show that
$$x\biggr(\sqrt{\frac{x}{x-1}}-1\biggr)= \frac{t}{\sqrt{t}+1}.$$
Can you proceed ?
A: $$x\sqrt{\frac{x}{x-1}}-x=\sqrt{\frac{x^3}{x-1}}-x=\sqrt{x^2+x+1+\frac{1}{x-1}}-x\\\approx \sqrt{x^2+x+\frac14}-x=\big(x+\frac12\big)-x=\frac12$$ 
You can rationalize that expression if you prefer, but I like the "completing the square" trick to just get it over with.
A: Hint:
Binomial expansion:
$\dfrac{x}{x-1}=1+\dfrac {1}{x-1}$
$(1+\dfrac{1}{x-1})^{1/2}=$
$1+(1/2)\dfrac{1}{x-1} +O((\dfrac{1}{x-1})^2)$
OR:
$y:=\sqrt{\dfrac{x}{x-1}};$ $y >0$;
$x= \dfrac{y^2}{y^2-1}$.
$\lim_{y \rightarrow 1}(\dfrac{y^2}{y^2-1})(y-1)=$
$\lim_{y \rightarrow 1}\dfrac{y^2}{y+1}=1/2.$
.
A: Hint:
$$
\sqrt{\frac{x}{x-1}} - 1 = \frac{\sqrt{x} - \sqrt{x-1}}{\sqrt{x-1}} = \frac{1}{\sqrt{x-1}(\sqrt{x} + \sqrt{x-1})} \sim \frac{1}{2x} \; (x \to \infty)
$$
A: By application of L' Hopital's rule: 
$$\lim_{x\to +\infty}\frac{\sqrt{\frac{x}{x-1}}-1}{\frac{1}{x}}=\left(\frac{0}{0}\right)=
\lim_{x\to +\infty}\frac{\left(\sqrt{\frac{x}{x-1}}-1\right)'}{\left(\frac{1}{x}\right)'}=
\ldots= \frac{1}{2}\lim_{x\to +\infty}\frac{\frac{x^2}{(x-1)^2}}{\sqrt{\frac{x}{x-1}}}=\frac{1}{2}\cdot 1=\frac12.$$
A: By binomial approximation 
$$\sqrt{\frac{x}{x-1}}= \sqrt{1+\frac{1}{x-1}}= 1+\frac{1}{2(x-1)}+o\left(\frac1x\right)$$
therefore 
$$\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)= \frac{x}{2(x-1)}+o\left(1\right)\to \frac12+0=\frac12$$
A: $$\lim_{x\to\infty}x\left(\sqrt{\frac{x}{x-1}}-1\right)=\lim_{x\to\infty}x\left(\left(1-\frac1x\right)^{-1/2}-1\right)=\lim_{y\to\infty}\frac{(1-y)^{-1/2}-1}{y}=\frac12.$$
A: 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*}
