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At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:

I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$: $$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt[3]{1+\cdots}+\sqrt{1+\cdots})}$$

II.) I tried to form $a^3-b^3$ in the numerator, same as I.): $$\frac{n^3(1-\sqrt[3]{1+\cdots})+5n^2+6}{n^2(\sqrt{1+\cdots}+\sqrt{1+\cdots}\sqrt{1+\cdots}+1)+3n+4}$$

What method can I apply here? Thank you.

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Are you familiar with Taylor series? – Clement C. Feb 3 '14 at 14:37
At first glance, the answer is $\infty$. – Alexander Feb 3 '14 at 14:38
use binomial theorem and binomial approximation,eg$(1+x)^n\approx 1+nx$ where $x<1$ – Half-Blood prince Feb 3 '14 at 14:39
up vote 2 down vote accepted

Hint: Combine the two approaches, $a^2-b^2$ and $a^3-b^3$. But calculate it as the limit of $$ (\root3\of{n^3+5n^2+6}-n)+(n-\sqrt{n^2+3n+4}), $$ and do the limits of the above two differences in parens separately.

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Try this:

$$\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} = n\cdot \left(\sqrt[3]{1+\frac{5}{n}+\frac{6}{n^3}} - \sqrt{1+\frac{3}{n}+\frac{4}{n^2}}\right).$$

L'Hospital rules should help from here on.

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Using series: \begin{align*} \sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} &= n\left( \left(1+\frac{5}{n}+\frac{6}{n^3}\right)^{1/3} - \left(1+\frac{3}{n}+\frac{4}{n^2}\right)^{1/2} \right) \\ &= n\left( 1+\frac{5}{3n} + o\left(\frac{1}{n}\right) - \left(1+\frac{3}{2n}+o\left(\frac{1}{n}\right)\right) \right) \\ &= n\cdot\frac{1}{6n} + o(1) \xrightarrow[n\to\infty]{}\frac{1}{6} \end{align*}

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i think there is some mistakes,we should get nswer as $\frac{1}{6}$ ,right? – Half-Blood prince Feb 3 '14 at 14:46
Yes (I had made a mistake at first, but fixed it). – Clement C. Feb 3 '14 at 14:47

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