# Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})$

Could you, please, check if I solved it right. \begin{align*} \lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) &= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)} - \sqrt[3]{n^3(1 + \frac1n)})\\ &= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\ &= \lim_{n \rightarrow \infty} (n - n) = 0. \end{align*}

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I do not think it is right: by the same reasoning, you would have lim(n+1-n) = lim(n(1+1/n)-n) = lim(n-n) = 0 – Jean Hominal Jul 1 '13 at 8:54
Sorry, it is not right. The result will be $1/6$. – André Nicolas Jul 1 '13 at 8:55
A free piece of advice. Before you try to prove the value a limit, it is often a good practice to calculate a few entries of the sequence. Here's homework for you: Plug in $n=10$, $n=100$, $n=1000,\ldots,$ to this formula with your pocket calculator. Observe the numbers that pop out. Does it look like they are closing in on a certain value? Don't overdo this, because with very large numbers the accuracy of your calculator will introduce errors to its output, when (as is the case here) nearly equal size numbers are subtracted from one another. – Jyrki Lahtonen Jul 1 '13 at 9:34

Here is a very calculus way to do the problem. (Your "taking" $n$ out is useful.) Write $h$ for $1/n$, and rewrite our limit as $$\lim_{h\to 0}\frac{(\sqrt{1+h} -1)-(\sqrt[3]{1+h}-1)}{h}, \tag{1}$$ because $$\frac{\sqrt{1+h} -1}{h} \quad\text{and}\qquad \frac{\sqrt[3]{1+h}-1}{h}$$ are familiar expressions whose limits we can compute. We recognize $$\lim_{h\to 0}\frac{\sqrt{1+h} -1}{h}$$ as the derivative of $\sqrt{1+x}$ at $x=0$, and $$\lim_{h\to 0}\frac{\sqrt[3]{1+h} -1}{h}$$ as the derivative of $\sqrt[3]{1+x}$ at $x=0$. These two derivatives, evaluated at $0$, are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. It follows that the limit (1) is equal to $\dfrac{1}{2}-\dfrac{1}{3}$.

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What comes after "So we want" is justified only if we know each of the separated limits exists finitely. By the work shown by the OP this might not be so clear to him/her. – DonAntonio Jul 1 '13 at 9:21
Thanks for the comment, I have reworded. In helping the OP, first priority would be to give a long list of concrete examples where a "substitution" algorithm doesn't work. – André Nicolas Jul 1 '13 at 11:28

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

$$=\lim_{n \rightarrow \infty} n\left(1+\frac1n\right)^{\frac12}-n\left(1+\frac1n\right)^{\frac13}$$

$$=\lim_{n \rightarrow \infty} n\left(1+\frac1{2n}+O\left(\frac1{n^2}\right)\right)-n\left(1+\frac1{3n}+O\left(\frac1{n^2}\right)\right)$$ (Using Taylor series or Generalized Binomial Theorem)

$$=\lim_{n \rightarrow \infty} \frac12+O\left(\frac1n\right)-\left(\frac13+O\left(\frac1n\right)\right)$$ $$=\frac12-\frac13$$

Alternatively, putting $h=\frac1n,$ $$(\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})=\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$

$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$ can be handled at least in three ways as follows:

$1:$ Using Taylor series,

$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h=\lim_{h\to0}\frac{1+\frac h2+O(h^2)-(1+\frac h3+O(h^2))}h$$ $$=\frac12-\frac13\text{ as }h\ne0\text{ as }h\to0$$

$2:$ $$\text{As }\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h \text{ is of the from }\frac\infty\infty,$$

Applying L'Hospital's Rule, $$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h=\lim_{h\to0}\left(\frac12\cdot(1+h)^{\frac12-1}-\frac1{3}\cdot(1+h)^{\frac13-1}\right)=\frac12-\frac13$$

$3:$ Like Jyrki Lahtonen has approached,

Putting $1+h=y^{\text{lcm}(2,3)}=y^6$

$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$

$$=\lim_{y\to1}\frac{y^3-y^2}{y^6-1}$$

$$=\lim_{y\to1}\frac{y^2(y-1)}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$$

$$=\lim_{y\to1}\frac{y^2}{y^5+y^4+y^3+y^2+y+1}\text{ as }y\ne1\text{ as } y\to1$$

$$= \frac{1^2}{1+1+1+1+1+1}$$

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+1 I quite like the substitution in your last approach (probably doesn't come as a surprise :-) – Jyrki Lahtonen Jul 1 '13 at 19:03

Others have spotted your error, and described ways of seeing that the limit is $1/6$. If you want to see a non-calculus way, then I suggest splitting it to two parts as follows: $$\sqrt{n^2+n}-\root3\of{n^3+n^2}=\left(\sqrt{n^2+n}-n\right)-\left(\root3\of{n^3+n^2}-n\right).$$ This operation is valid, if you can show that both sequences on the r.h.s converge. To see that I would interpret $n=\sqrt{n^2}$ in the first summand and $n=\root3\of{n^3}$ in the second. Then use the usual tricks $$\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b},$$ and $$\root 3\of a-\root3\of b=\frac{a-b}{a^{2/3}+(ab)^{1/3}+b^{2/3}}.$$ These come from the respective polynomial factorizations $$x^2-y^2=(x-y)(x+y),\qquad x^3-y^3=(x-y)(x^2+xy+y^2).$$

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Slick! $\textbf{}$ – Potato Jul 1 '13 at 17:23

$$(\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) = \sqrt{n^2\left(1 + \frac1n\right)} - \sqrt[3]{n^3\left(1 + \frac1n\right)}=$$

$$=n\left(\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}\right)=\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}$$

Looking at the above as function of the continuous variable $\,n\,$, we can use l'Hospital to try to calculate its limit:

$$\lim_{n\to\infty}\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}\stackrel{\text{l'H}}=\lim_{n\to\infty}\left(\frac1{2\sqrt{1+\frac1n}}-\frac1{3\sqrt[3]{\left(1+\frac1n\right)^2}}\right)=\frac12-\frac13=\frac16$$

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You made a mistake when you applied the limit $n \to \infty$ to the $\left(1 + \frac 1n\right)$ terms but not other terms outside. This is not a valid limit application because the whole thing is still in an indeterminate $\infty - \infty$ form.

I can offer you a tedious method that does not require you to estimate the growth of the two terms in the difference. (In fact, if you expand out everything in $O(\ldots)$, you only need basic algebra and some fundamental limit theorems.)

Define \begin{align*} A_n & = \sqrt{n^2 + n} = \sqrt[6]{(n^2 + n)^3} = \sqrt[6]{n^6 + 3n^5 + 3n^4 + n^3} = n\sqrt[6]{1 + O(1/n)}\\ B_n & = \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^3 + n^2)^2} = \sqrt[6]{n^6 + 2n^5 + n^4} = n\sqrt[6]{1 + O(1/n)}. \end{align*} We want to find $\lim_{n\to\infty} A_n - B_n$. Let us factor $A_n^6 - B_n^6$ as follows: \begin{align*} A_n^6 - B_n^6 & = (A_n - B_n)(A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5) \\ \therefore A_n - B_n & = \frac{A_n^6 - B_n^6}{A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5} \end{align*} We know that $A_n^6 - B_n^6 = n^5 + 2n^4 + n^3 = n^5(1 + O(1/n))$. All terms in the denominator are of the form $A^i_nB^{5-i}_n$, so $$A^i_nB^{5-i}_n = \sqrt[6]{(n^6 + 3n^5 + 3n^4 + n^3)^i(n^6 + 2n^5 + n^4)^{5-i}} = n^5\sqrt[6]{1 + O(1/n)}.$$ Therefore, \begin{align*} \lim_{n\to\infty} A_n - B_n & = \lim_{n\to\infty}\frac{n^5(1 + O(1/n))}{6n^5\sqrt[6]{1 + O(1/n)}} \\ & = \frac 16 \end{align*}

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Forget your last term and Just set x = 1/n in the second

The first term of the series is x/2 - x/3=x/6. Since x=1/n, the result is just 1/6

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Although it seems like a pain, you can multiply by "the conjugate" here, too. But you want $a^6-b^6$ on too, so multiply by $a^5+a^4b+...+b^5$ on top and bottom. The too cancels out nicely. The bottom seems bad, but remember that every term on the bottom grows like $n^2$, and there are six terms, thus giving 1/6.

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