Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
