# Limit of $\sqrt{x^3/(x-3)}-x$ as $x\to\infty$

I typed the following expression into Wolfram: sqrt[(x^3)/(x-3)]-x and wanted to know the limit for $x$ approaching infinity. The result is $3/2$.

I have spent a lot of time now trying to get $3/2$ on paper by myself but I just cannot get to the right result.

I'm not subscribed to Wolfram so I cannot see their step by step solution so if anybody could help out here, I'd be much obliged.

Thanks

Hint: For $x>3$, using identity $(a+b)(a-b)=a^2-b^2$ $$\sqrt{\dfrac{x^3}{x-3}}-x=\dfrac{\frac{x^3}{x-3}-x^2}{\sqrt{\frac{x^3}{x-3}}+x}$$

$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x-3}}-x\right)= \lim_{x\to\infty}x\frac{\sqrt{x}-\sqrt{x-3}}{\sqrt{x-3}}= \lim_{x\to\infty}\frac{3x}{\sqrt{x-3}(\sqrt{x}+\sqrt{x-3})}$$ Divide numerator and denominator by $x$.

HINT:

Set $1/x=h\implies h\to0^+, h>0\implies\sqrt{h^2}=|h|=+h$

to find

$$\lim_{h\to0^+}\left(\dfrac1{h\sqrt{1-3h}}-\dfrac1h\right) =\lim_{h\to0^+}\dfrac1{\sqrt{1-3h}}\cdot\lim_{h\to0^+}\dfrac{1-\sqrt{1-3h}}h$$

Now rationalize the numerator and use the fact $:h\to0\implies h\ne0$

You have been given several good solutions to the problem.

The first think I would have done is to set $x=\frac 1h$; so $$\frac{x^3}{x-3}=\frac 1 {h^2-3h^3}=\frac 1 {h^2} \frac 1 {1-3h}$$ $$\sqrt{\frac{x^3}{x-3}}=\frac 1h\frac 1{\sqrt{1-3h}}$$ Now, using the generalized binomial theorem $$(1-3h)^{-1/2}=1+\frac{3 h}{2}+\frac{27 h^2}{8}+\cdots$$ $$\frac 1h\frac 1{\sqrt{1-3h}}=\frac 1 h \Big(1+\frac{3 h}{2}+\frac{27 h^2}{8}+\cdots)=\frac 1h+\frac{3 }{2}+\frac{27 h}{8}+\cdots$$ and finally $$\frac 1h\frac 1{\sqrt{1-3h}}-\frac 1 h=\frac{3 }{2}+\frac{27 h}{8}+\cdots$$ which shows the limit and also how it is approached?

• Thanks everyone for replies, I'd like to add that I'm trying to solve this using l'Hopitals where I get a fraction and solve the derivative of the numerator and denominator but what I'm getting is 3/2-9x instead of 3/2. Feb 29, 2016 at 14:19