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$$\lim_{n \to \infty} \frac{(\sqrt{n}+3n)^2-9n^2}{n^{3/2}+7n}$$

I'm stuck and would appreciate some assistance :D

Here's what I did so far: $$ \begin{split} \lim_{n \to \infty} \frac{(\sqrt{n}+3n)^2-9n^2}{n^\frac{3}{2}+7n} &= \lim_{n \to \infty} \frac{n+6n \cdot \sqrt{n} + 9n^2 - 9n^2}{n^\frac{3}{2}+7n} \\ &= \lim_{n \to \infty} \frac{n+6n \cdot \sqrt{n}}{n^{3/2}+7n} \end{split} $$

we divide by $n$ :

$$\lim_{n \to \infty} \frac{(\sqrt{n}+3n)^2-9n^2}{n^\frac{3}{2}+7n} =\lim_{n \to \infty} \frac{1+6 \cdot n^{\frac{1}{2}}}{7 + n^\frac{1}{2}} = \lim_{n \to \infty} \frac{1+6 \cdot \sqrt{n}}{7 + \sqrt{n}}$$

At this point I'm stuck. The limit should be 6. Any suggestions?

Comment said dividing by $\sqrt{n}$, so here we go :D

$$\lim_{n \to \infty} \left( \frac{(\sqrt{n}+3n)^2-9n^2}{n^\frac{3}{2}+7n}\right) =\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n}}+6}{\frac{7}{\sqrt{n}} + 1} = \frac{6}{1} = 6.$$

Okay got it thanks :D

Arturo Magidin
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jenny
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    Now try dividing by $\sqrt{n}$ :) – Alann Rosas May 12 '21 at 17:42
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    basic rule of thumb on limits of ratios with sums in them, only the fastest growing term matters. Thus the 1 and 7 can be "ignored". To do it formally, you divide by the fastest growing term and all the slower ones go to 0 – Alan May 12 '21 at 17:45

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