# Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$

Please help me to calculate the following limit $$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$ I factored out $$n$$ from both radicals but it didn't work and tried to use the identity $$a^2-b^2$$ and $$a^3-b^3$$.

• What have you tried? – Robert Shore May 21 at 0:19
• I factored out n from both radicals but it didn't work and tried to use the identity a^2-b^2 and a^3-b^3. – Andrei Umbriță May 21 at 0:21
• Please edit your question to indicate your attempts (rather than putting that information in the comments). Your question is much more likely to get a positive reception if you do so. We are sensitive to people simply showing up and asking a question without any indication they've put effort into finding the solution for themselves. – Robert Shore May 21 at 0:23
• are you familiar with binomial series? – J. W. Tanner May 21 at 0:27
• Yes, but I would like to see an answer without these or approximations. – Andrei Umbriță May 21 at 0:28

For this type of problem, my preferred weapon is the generalized binomial theorem in the form $$(1+x)^a =1+ax+a(a-1)x^2/2+O(x^3)$$.

Often the simpler form $$(1+x)^a =1+ax+O(x^2)$$ is enough.

So,

$$\begin{array}\\ \sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} &=n\sqrt[3]{1+3/n}-n\sqrt{1+2/n}\\ &=n(1+(3/n)(1/3)+O(1/n^2))-n(1+(2/n)(1/2)+O(1/n^2))\\ &=n(1+1/n+O(1/n^2))-n(1+1/n+O(1/n^2))\\ &=n+1+O(1/n)-n-1-O(1/n)\\ &=O(1/n)\\ &\to 0 \qquad\text{as } n \to \infty\\ \end{array}$$

Note that if what was wanted was $$n$$ times the difference, the additional term in the expansion would have been needed.

You might try this for practice.

• by "difference", did you mean the case in Riemann's answer? is that rigorous? – farruhota May 21 at 6:29
• Yes and yes. The additional term will give the 1/2n result. – marty cohen May 21 at 15:10
• without the additional term it is indeterminate $0/0$, isn't it? Thank you. +1 – farruhota May 21 at 15:24
• No. It comes to O(1). – marty cohen May 21 at 16:43
• which is not demonstrated, $=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}}$ with $\sqrt[3]{1+\frac{3}{n}}-1 \sim \frac{1}{n},\sqrt{1+\frac{2}{n}}-1\sim \frac{1}{n}$ are not sufficient, aren't they? – farruhota May 21 at 17:50

Another way to solve the problem is the following, where no Differential Calculus is needed.

By elementary algebraic manipulation of radicals, you get:

$$\begin{split} \sqrt[3]{n^3 + 3n^2} - \sqrt{n^2 + 2n} &= \underbrace{\sqrt[6]{(n^3 + 3n^2)^2}}_{=:\sqrt[6]{a}} - \underbrace{\sqrt[6]{(n^2 + 2n)^3}}_{=:\sqrt[6]{b}} \\ & = \frac{a - b}{\sqrt[6]{a^5} + \sqrt[6]{a^4b} + \sqrt[6]{a^3b^2} + \sqrt[6]{a^2b^3} + \sqrt[6]{ab^4} + \sqrt[6]{b^5}} \\ &\approx \frac{-3n^4}{6n^5} \\ &= - \frac{1}{2n} \end{split}$$

with the latter member going to $$0$$ as $$n \to \infty$$.

First let's get rid of the square root:

$$\lim_{n \to \infty} (\sqrt[3]{n^3-3n^2}-\sqrt{n^2+2n})=\lim_{n \to \infty} \frac{\sqrt[3]{(n^3+3n^2)^2}-(n^2+2n)}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}\\= -\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}.$$

Now let's get rid of the cube root:

$$-\lim_{n \to \infty} \frac{n^2+2n-\sqrt[3]{(n^3+3n^2)^2}}{\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n}}$$

$$=-\lim_{n \to \infty} \frac{(n^2+2n)^3-(n^3+3n^2)^2}{(\sqrt[3]{(n^3+3n^2)}+\sqrt{n^2+2n})((n^2+2n)^2+(n^2+2n)\sqrt[3]{(n^3+3n^2)^2}+\sqrt[3]{(n^3+3n^2)^4})}.$$

The numerator yields $$-3n^4$$ and a bunch of lower-order terms, and the denominator yields $$6n^5$$ and a bunch of lower order terms, so dividing numerator and denominator by $$n^5$$ shows that the answer is $$0$$.

Use the following fact: when $$x\to 0$$, $$(1+x)^{\alpha}-1\sim\alpha x.$$ So $$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}=n\left(\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)\right)$$ $$=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}},$$ use the above fact: $$\sqrt[3]{1+\frac{3}{n}}-1 \sim \frac{1}{n},\sqrt{1+\frac{2}{n}}-1\sim \frac{1}{n},$$ and this implies $$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )=0.$$

In fact: $$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}\sim -\frac{1}{2n},n\to\infty.$$

For positive $$n$$, let $$a=\sqrt[3]{n^3+3n^2}$$, and let $$b=\sqrt{n^2+2n}$$.

Note that $$a < n+1$$ and \begin{align*} 3n +1 &= (n+1)^3-a^3\\[4pt] &= \bigl((n+1)-a\bigr)\bigl((n+1)^2+a(n+1)+a^2)\\[4pt] &> \bigl((n+1)-a\bigr)(n+1)^2\\[4pt] \end{align*} hence $$0 < n+1-a < {\large{\frac{3n+1}{(n+1)^2}}}$$, which implies $$\displaystyle{\lim_{n\to \infty}}\;(n+1)-a = 0$$.

Also, $$b < n+1$$ and \begin{align*} 1&=(n+1)^2-b^2\\[4pt] &= \bigl((n+1)-b\bigr)\bigl((n+1)+b\bigr)\\[4pt] &> \bigl((n+1)-b\bigr)(n+1)\\[4pt] \end{align*} hence $$0 < (n+1)-b < {\large{\frac{1}{n+1}}}$$, which implies $$\displaystyle{\lim_{n\to \infty}}\;(n+1)-b = 0$$.

Then we get $$\lim_{n\to \infty}\;a-b = \lim_{n\to \infty}\;\bigl(a - (n+1)\bigr) + \bigl((n+1)-b\bigr) = 0 + 0 = 0$$

• The question has changed. – Nosrati May 21 at 0:52
• Ok, I edited my answer to match your edited question. – quasi May 21 at 1:01

There is a way to use directly the two binomial formulas you mentioned. It only needs a little trick:

$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} = \sqrt[3]{n^3+3n^2}\color{blue}{-n + n}-\sqrt{n^2+2n}$$ Now, consider $$\begin{eqnarray*} \sqrt[3]{n^3+3n^2}-n & \stackrel{n=\sqrt[3]{n^3}}{=} & \frac{n^3+3n^2 - n^3}{(\sqrt[3]{n^3+3n^2})^2 + n\sqrt[3]{n^3+3n^2}+ n^2}\\& = & \frac{3}{\left(\sqrt[3]{1+\frac{3}{n}}\right)^2 + \sqrt[3]{1+\frac{3}{n}} + 1}\\ & \stackrel{n \to \infty}{\longrightarrow} & 1 \end{eqnarray*}$$

Similarly, you get $$n - \sqrt{n^2+2n} \stackrel{n \to \infty}{\longrightarrow} -1$$.

Hence, you get

$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} ) = \lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-n ) + \lim_{n \to \infty} (n-\sqrt{n^2+2n} ) = 1-1 = 0$$

Set $$1/x=h,h\to0^+$$

The limit becomes $$\lim_{h\to0}\dfrac{\sqrt[3]{1+3h}-1}h-\lim...\dfrac{\sqrt{1+2h}-1}h$$

Now for $$F=\lim_{...}\dfrac{\sqrt[m]{1+mh}-1}h,$$

set $$\sqrt[m]{1+mh}-1=u,$$

$$\implies mh=(1+u)^m-1=mu+O(u^2)$$

Can you take it from here?