Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series.

Question

$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$

My try is as follows:

$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to ∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to ∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$ which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?

How does generalised binomial theorem make the limit difficult? You would just get$$x\left(1+\frac1x+o\left(\frac1x\right)-\left(1-\frac1x+o\left(\frac1x\right)\right)\right)=2+o(1)$$hence the limit is $2$. — Peter Foreman, Oct 25 '19 at 12:21
thanks, I understand your solution, but is there any elementary way to solve the problem? — Absurd, Oct 25 '19 at 12:24

Robert Z · Accepted Answer · 2019-10-25T12:55:22.380

We first note that for any positive integer $n$ and any real $a$, $$\lim_{x\to \infty}x\left(\sqrt[n]{1+\frac{a}{x}}-1\right)= \lim_{s\to 1}a\frac{s-1}{s^n-1}=\lim_{s\to 1}\frac{a}{s^{n-1}+s^{n-2}+\dots +s +1}=\frac{a}{n}$$ where $s=\sqrt[n]{1+a/x}$ and therefore $a/x=s^n-1$, and $x=a/(s^n-1)$.

Hence, from your work, we split the limit in two: $$\begin{align}\lim_{x\to +\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}) &=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right) \\&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-1\right)-\lim_{x\to \infty}x\left(\sqrt[2]{1 +\frac{-2}{x}}-1\right)\\&=\frac{3}{3}-\frac{-2}{2}=1+1=2. \end{align}$$

P.S. Note that on the other side, $$\lim_{x\to -\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=-\infty$$

,awesome, I just think you forgot an $a$ in the third line. – Absurd Oct 25 '19 at 12:38 — Absurd, Oct 25 '19 at 12:38
it's right now,thanks for the great answer. – Absurd Oct 25 '19 at 12:45 — Absurd, Oct 25 '19 at 12:45

score 2 · Answer 2 · answered Oct 25 '19 at 12:53

Using high-school math:$$\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=\lim_{x\to ∞} \frac{\sqrt[3]{(x^{3}+3x^{2})^2}-x^{2}+2x}{\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x}}=$$ $$ \lim_{x\to ∞} \frac{(x^{3}+3x^{2})^2-(x^{2}-2x)^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$\lim_{x\to ∞} \frac{12x^5-3x^4+8x^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$[\text{leaving the highest power}]=\lim_{x\to ∞} \frac{12x^5}{(x+x)(x^4+x^4+x^4)}=2$$

score 1 · Answer 3 · answered Oct 25 '19 at 12:50

A quite elementary way is just using the two binomial formulas $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$ as follows:

\begin{eqnarray*}\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x} & = & (\sqrt[3]{x^{3}+3x^{2}} - x) + (x -\sqrt{x^{2}-2x})\\ & = & \frac{3x^2}{\sqrt[3]{(x^{3}+3x^{2})^2} + x\sqrt[3]{x^{3}+3x^{2}} + x^2} + \frac{2x}{x+\sqrt{x^{2}-2x}}\\ & = & \frac{3}{\sqrt[3]{(1+\frac{3}{x})^2} + \sqrt[3]{1+\frac{3}{x}} + 1} + \frac{2}{1+\sqrt{1-\frac{2}{x}}} \\ & \stackrel{x\to \infty}{\longrightarrow} & 1+1 = 2 \end{eqnarray*}

score 1 · Answer 4 · answered Oct 25 '19 at 13:00

1

By binomial approximation

$\sqrt[3]{x^{3}+3x^{2}}=x\sqrt[3]{1+3/x}\approx x\left(1+\frac1x\right)=x+1$
$\sqrt{x^{2}-2x}=x\sqrt{1-2/x}\approx x\left(1-\frac1x\right)=x-1$

therefore

$$\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}\approx 2$$

answered Oct 25 '19 at 13:00

user

154,566

score 0 · Answer 5 · answered Oct 25 '19 at 12:44

Hint:

Set $1/n=h$

$$x^3+3x^2=\dfrac{1+3h}{h^3}, x^2-2x=\dfrac{1-2h}{h^2}$$

$$\large \lim_{x\to\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$ $$=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+3h}-\sqrt{1-2h}}h$$ $$=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+3h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt{1-2h}-1}h$$

Set $\sqrt[3]{1+3h}-1=p\implies1+3h=(1+p)^3, h=p+p^2+\dfrac{p^3}3$

and $\sqrt{1-2h}-1=q\implies1-2h=(1+q)^2,-h=q+\dfrac{q^2}2$

score -1 · Answer 6 · answered Oct 25 '19 at 12:30

From OP's work $$L=\lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{x} \right)^{1/3}-\left( 1-\frac{2}{x} \right)^{1/2}\right]= \lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{3x} \right)-\left( 1-\frac{2}{2x} \right)\right]= \lim_{x \rightarrow \infty}x \frac{2}{x}=2. $$

Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series.

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