Limit , need tips please

Question

Limit of $(n^2+n)^{1/2} - (n^3+n^2)^{1/3}$ as $n\to\infty$

Help me please solving this.

Answer 1

The approach is standard, using the fact that $a-b = \dfrac {a^k - b^k} {a^{k-1} + a^{k-2}b + \dots + a b^{k-2} + b^{k-1}}$. In our case, $a = \sqrt{n^2 + n} = \sqrt[6]{(n^2 + n)^3}$ and $b = \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^3 + n^2)^2}$, so we shall take $k=6$. It follows that

$$\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^2 + n)^3} - \sqrt[6]{(n^3 + n^2)^2} = \frac {(n^2 + n)^3 - (n^3 + n^2)^2} {\sum \limits _{i=1} ^6 \sqrt{n^2 + n}^{6-i} \sqrt[3]{n^3 + n^2}^{i-1}} = \\ \frac {(n^6 + 3n^5 + 3n^4 + n^3) - (n^6 + 2n^5 + n^4)} {n^5 \sum \limits _{i=1} ^6 \sqrt{1 + \frac 1 n}^{6-i} \sqrt[3]{1 + \frac 1 n}^{i-1}} = \frac {1 + 2 \frac 1 n + \frac 1 {n^2}} {\sum \limits _{i=1} ^6 \sqrt{1 + \frac 1 n}^{6-i} \sqrt[3]{1 + \frac 1 n}^{i-1}} \to \frac {1 + 0 + 0} {\sum \limits _{i=1} ^6 1 \cdot 1} = \color{red} {\frac 1 6} .$$

Answer 2

We may apply binomial expansion to see that

$$(n^2+n)^{1/2}=n+\frac12+\mathcal O(n^{-1})$$

$$(n^3+n^2)^{1/3}=n+\frac13+\mathcal O(n^{-1})$$

So as $n\to\infty$,

$$(n^2+n)^{1/2}-(n^3+n^2)^{1/3}\sim\color{red}{\frac16}+\mathcal O(n^{-1})$$

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For explanation, we have the binomial expansion:

$$(a+b)^k=a^k+ka^{k-1}b+\mathcal O(a^{k-2}b^2)$$

If you look at your regular old binomial expansion, the first number is obviously $a^k$ and the second is always the exponent times $a^{k-1}b$, and it goes on with different coefficients on and on. However, its a little harder figuring out the coefficients beyond this point, so we just sum it all up with $\mathcal O(a^{k-2}b^2)$, which basically says it is equal to that multiplied by some finite constant. Terms beyond that are even smaller than $a^{k-2}b^2$, so they get combined into it all.

For example,

$$(n^2+n)^{1/2}=(n^2)^{1/2}+\frac12(n^2)^{-1/2}n+\mathcal O((n^2)^{-3/2}n^2)\\=n+\frac12+\mathcal O(n^{-1})$$

Notice that regardless of constant, as $n\to\infty$, $\mathcal O(n^{-1})\to0$, so this method is very simple and effective for these problems.

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You could alternatively note that

$$(n^2+n)^{1/2}-(n^3+n^2)^{1/3}=\frac{(1+x)^{1/2}-1}x-\frac{(1+x)^{1/3}-1}x$$

where $x=1/u$. As $u\to\infty$, $x\to0$, so

$$\lim_{x\to0}\frac{(1+x)^{1/2}-1}x-\frac{(1+x)^{1/3}-1}x=\left.\frac d{dx}(1+x)^{1/2}-(1+x)^{1/3}\right|_{x=0}$$