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

Question

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 $a^3-b^3$.

Answer 1

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

Answer 2

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 $0$.

Answer 3

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.

Answer 4

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 $n \to \infty$.

Answer 5

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

Answer 6

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

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

Answer 7

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?