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I am supposed to find the limit of this sequence. In my lecture notes I am told to always take out the fastest growing term in the expression and divide by this term. e.g if I have n/n+1 the fastest growing term is n and I take this out from the numerator and the denominator. In this expression the fastest growing term is sqrrt(3n^3+2n) but they have divided this by n^3/2 and I don't understand why when that is not the fastest growing term. Unless I treat sqrrt(3n^3+2n) not as one single term but two sepearte terms of 3n^3 and 2^n but surely thats not right and sqrrt(3n^3+2n) is one single term right?

$$\frac{\sqrt{3n^3+2n} +\sqrt[\large 3]{2n^2+3n}}{n\sqrt{2n}+n+\sqrt{2n}}$$

user134785
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2 Answers2

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Notice that $3n^3$ is more faster than $2n$ so $$\sqrt{3n^3+2n}\sim_\infty \sqrt3 n^{3/2}$$ and by the same method we have $$\sqrt[3]{2n^2+3n}\sim_\infty\sqrt[3]{2}n^{2/3}$$ and comparing the two last terms we find that the numerator is symptomatically equivalent to $\sqrt3 n^{3/2}$. Do the same method for the denominator and we find

$$\frac{\sqrt{3n^3+2n} +\sqrt[\large 3]{2n^2+3n}}{n\sqrt{2n}+n+\sqrt{2n}}\sim_\infty\frac{\sqrt3 n^{3/2}}{\sqrt2 n^{3/2}}=\sqrt{\frac32}$$

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Rewrite each term in the numerator, and rewrite each term in the denominator, so that you have a power of $n$ times things that are either constant or approach $0.$

$$\frac{\sqrt{3n^3+2n} \; + \; \sqrt[\large3]{2n^2+3n}}{n\sqrt{2n} \; + \; n \; + \; \sqrt{2n}} \;\; = \;\; \frac{\sqrt{n^3\left(3 + \frac{2}{n^2}\right)} \; + \; \sqrt[\large3]{n^2\left(2 + \frac{3}{n}\right)}}{n\sqrt{2n} \; + \; n \; + \; \sqrt{2n}} $$

$$ = \;\; \frac{n^{\frac{3}{2}}\sqrt{3 + \frac{2}{n^2}} \;\; + \;\; n^{\frac{2}{3}}\sqrt[\large3]{2 + \frac{3}{n}}}{n^{\frac{3}{2}}\sqrt{2} \; + \; n^1 \; + \; n^{\frac{1}{2}}\sqrt{2}}$$

The highest of these powers of $n$ is $n^{\frac{3}{2}},$ so we divide numerator and denominator by $n^{\frac{3}{2}},$ or equivalently, we multiply numerator and denominator by $n^{-\frac{3}{2}}.$

$$ \frac{n^{\frac{3}{2}}\sqrt{3 + \frac{2}{n^2}} \;\; + \;\; n^{\frac{2}{3}}\sqrt[\large3]{2 + \frac{3}{n}}}{n^{\frac{3}{2}}\sqrt{2} \; + \; n^1 \; + \; n^{\frac{1}{2}}\sqrt{2}} \; \cdot \; \frac{n^{ -\frac{3}{2} }} {n^{-\frac{3}{2}}}$$

$$ = \;\; \frac{n^{-\frac{3}{2}} \cdot n^{\frac{3}{2}}\sqrt{3 + \frac{2}{n^2}} \;\; + \;\; n^{-\frac{3}{2}} \cdot n^{\frac{2}{3}}\sqrt[\large3]{2 + \frac{3}{n}}}{n^{-\frac{3}{2}} \cdot n^{\frac{3}{2}}\sqrt{2} \; + \; n^{-\frac{3}{2}} \cdot n^1 \; + \; n^{-\frac{3}{2}} \cdot n^{\frac{1}{2}}\sqrt{2}}$$

$$ = \;\; \frac{\sqrt{3 + \frac{2}{n^2}} \;\; + \;\; n^{-\frac{5}{6}}\sqrt[\large3]{2 + \frac{3}{n}}}{\sqrt{2} \; + \; n^{-\frac{1}{2}} \; + \; n^{-1}\sqrt{2}} \;\; \longrightarrow \;\; \frac{\sqrt{3 + 0} \;\; + \;\; 0 \cdot \sqrt[\large3]{2 + 0}}{\sqrt{2} \; + \; 0 \; + \; 0 \cdot \sqrt{2}} \;\; = \;\; \frac{\sqrt{3}}{\sqrt{2}}$$

Dave L. Renfro
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