# 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.

• I thought you can use L'Hôpitals only when the limit evaluates to $\frac{\infty}{\infty}$ or $\frac{0}{0}$. – Radu Gabriel Nov 7 at 10:18
• Yes, but you can rewrite the term as $\frac{\sqrt{\frac{x}{x-1}} -1}{\frac{1}{x}}$, where both numerator and denominator go to 0. – Viktor Glombik Nov 7 at 10:22
• Oh... I was gonna say, it seems to me that this is always the case when a limit evaluates to $\frac{\infty}{0}$. I looked at ways to rewrite it before I decided to ask here, sometimes I just can't see it. Thanks a lot though. – Radu Gabriel Nov 7 at 10:24
• Does $x$ tend to $+\infty$ or $-\infty$? – Bernard Nov 7 at 10:34
• $+\infty$, otherwise I would have specified. But the limit would go to $\frac{1}{2}$ either way. – Radu Gabriel Nov 7 at 10:42

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 ?

$$x\sqrt{\frac{x}{x-1}}-x=\sqrt{\frac{x^3}{x-1}}-x=\sqrt{x^2+x+1+\frac{1}{x-1}}-x\\\approx \sqrt{x^2+x+\frac14}-x=\big(x+\frac12\big)-x=\frac12$$

You can rationalize that expression if you prefer, but I like the "completing the square" trick to just get it over with.

• I do not understand how do you go from $\sqrt{x^2+x+1+\frac{1}{x-1}}-x$ to $\sqrt{x^2+x+\frac14}-x$. – Radu Gabriel Nov 7 at 10:44
• @RaduGabriel As $x\to\infty$, the last two terms under the radical are negligible. They could be removed, but they could also be substituted with another negligible term that would complete the square. – Matthew Daly Nov 7 at 10:55
• I did not know you can do that, thanks a lot. – Radu Gabriel Nov 7 at 10:57
• @RaduGabriel Your teacher probably doesn't either. :P math.stackexchange.com/questions/3425688/… provides some justification for the technique. – Matthew Daly Nov 7 at 11:53

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:

$$y:=\sqrt{\dfrac{x}{x-1}};$$ $$y >0$$;

$$x= \dfrac{y^2}{y^2-1}$$.

$$\lim_{y \rightarrow 1}(\dfrac{y^2}{y^2-1})(y-1)=$$

$$\lim_{y \rightarrow 1}\dfrac{y^2}{y+1}=1/2.$$

.

Hint: $$\sqrt{\frac{x}{x-1}} - 1 = \frac{\sqrt{x} - \sqrt{x-1}}{\sqrt{x-1}} = \frac{1}{\sqrt{x-1}(\sqrt{x} + \sqrt{x-1})} \sim \frac{1}{2x} \; (x \to \infty)$$

By application of L' Hopital's rule:

$$\lim_{x\to +\infty}\frac{\sqrt{\frac{x}{x-1}}-1}{\frac{1}{x}}=\left(\frac{0}{0}\right)= \lim_{x\to +\infty}\frac{\left(\sqrt{\frac{x}{x-1}}-1\right)'}{\left(\frac{1}{x}\right)'}= \ldots= \frac{1}{2}\lim_{x\to +\infty}\frac{\frac{x^2}{(x-1)^2}}{\sqrt{\frac{x}{x-1}}}=\frac{1}{2}\cdot 1=\frac12.$$

$$\lim_{x\to\infty}x\left(\sqrt{\frac{x}{x-1}}-1\right)=\lim_{x\to\infty}x\left(\left(1-\frac1x\right)^{-1/2}-1\right)=\lim_{y\to\infty}\frac{(1-y)^{-1/2}-1}{y}=\frac12.$$

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

$$\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)= \frac{x}{2(x-1)}+o\left(1\right)\to \frac12+0=\frac12$$

Her is an elementary way:

$$\begin{eqnarray*} \biggr(x\sqrt{\frac{x}{x-1}}-x\biggr) & = & x \biggr(\sqrt{\frac{x}{x-1}}-1\biggr) \\ & = & x \biggr(\frac{\frac{x}{x-1}-1}{\sqrt{\frac{x}{x-1}}+1}\biggr) \\ & = & x \biggr(\frac{1}{(x-1)\sqrt{\frac{x}{x-1}}+(x-1)}\biggr) \\ & = & \frac{1}{(1-\frac{1}{x})\sqrt{1+\frac{1}{x-1}}+1-\frac{1}{x}} \\ & \stackrel{x \to \infty}{\longrightarrow} & \frac{1}{(1-0)\sqrt{1+0}+1-0} = \frac{1}{2} \end{eqnarray*}$$