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Prove by definition the following limit: $$\lim_{n\rightarrow \infty}\sqrt{\frac{2n^3+3n-1}{8n^3+n^2}}=\frac{1}{2}$$

For the first i've been tried rationalize: $$\begin{align}\left|\frac{\sqrt{2n^3+3n-1}}{\sqrt{8n^3+n^2}}-\frac{1}{2}\right|&=\left|\frac{2\sqrt{2n^3+3n-1}-\sqrt{8n^3+n^2}}{2\sqrt{8n^3+n^2}}\right|\\&=\left|\frac{4(2n^3+3n-1)-(8n^3+n^2)}{2\sqrt{8n^3+n^2}\left(2\sqrt{2n^3+3n-1}+\sqrt{8n^3+n^2}\right)}\right|\\&\leq \left|\frac{-n^2+12n-4}{2\sqrt{8n^3+n^2}\left(2\sqrt{2n^3+3n-1}+\sqrt{8n^3+n^2}\right)}\right|\end{align}$$

but then, I don't know how to proceed.

The second limit is: $$\lim_{n\rightarrow \infty} \frac{n!}{x^n}=\infty\qquad (x>0)$$ I can't use the reciprocal limit, and I won't see how use squezze theorem or Archmidean property... any hint? Or book?

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    $\begingroup$ Hint: Show if $\lim f(n)=\frac{1}{4}$ then $\lim \sqrt{f(n)}=\frac{1}{2}.$ Don't get bogged down in the details of what $f(n)$ is initially. $\endgroup$ – Thomas Andrews Apr 4 at 16:21
  • $\begingroup$ I know the result, but the instructions said "direct from definition", that's my problem $\endgroup$ – Ragnar1204 Apr 4 at 16:26
  • $\begingroup$ Well "or some technique" is a little vague. $\endgroup$ – Thomas Andrews Apr 4 at 16:32
  • $\begingroup$ Yes I know, but the "some technique" is for the second limit. Sorry $\endgroup$ – Ragnar1204 Apr 4 at 16:50
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$$\begin{align}\left|\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}-\frac{1}{2}\right| &=\left|\frac{{\frac{2n^3+3n-1}{{8n^3+n^2}}}-\frac{1}{4}}{\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}+\frac{1}{2}}\right|\\ &=\left|\frac{\frac{-n^2+12n-4}{32n^3+4n^2}}{\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}+\frac{1}{2}}\right|\\ &<\left|\frac{\frac{-n^2+12n-4}{32n^3+4n^2}}{\frac{1}{2}}\right|\\ &=2\left|\frac{1-12\frac{1}{n}+4\frac{1}{n^2}}{32n+4}\right|\\ &<\frac{1}{16n+2} \end{align}$$

for $n$ large (say $n>12.$)


This is the long way of showing that $f(n)=\frac{2n^3+3n-1}{8n^3+n^2}$ satisfies $f(n)\to \frac{1}{4}$ since:

$$\left|f(n)-\frac{1}{4}\right|=\left|\frac{1-12\frac{1}{n}+4\frac{1}{n^2}}{32n+4}\right|<\frac{1}{32n+4}$$ for $n>12$, and then:

$$\left|\sqrt{f(n)}-\frac{1}{2}\right|=\frac{\left|f(n)-\frac{1}{4}\right|}{\sqrt{f(n)}+\frac{1}{2}}<2\left|f(n)-\frac14\right|$$

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$$\sqrt{\frac{2n^3+3n-1}{8n^3+n^2}} =\sqrt{\frac{2+\frac{3}{n^2}-\frac{1}{n^3}}{8+\frac{1}{n}}} \rightarrow \sqrt{\frac{2}{8}} = \frac{1}{2} $$

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    $\begingroup$ That's fine if you know $\sqrt{\cdot}$ is continuous, but the question does say "from definition." so you need to show both that $f(n)\to\frac{1}{4}$ and from there show that $\sqrt{f(n)}\to\frac{1}{2}$ from definition. $\endgroup$ – Thomas Andrews Apr 4 at 16:34

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