Some limits of sequences

Question

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?

Answer 1

$$\begin{align}\left|\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}-\frac{1}{2}\right| &=\left|\frac{{\frac{2n^3+3n-1}{{8n^3+n^2}}}-\frac{1}{4}}{\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}+\frac{1}{2}}\right|\\ &=\left|\frac{\frac{-n^2+12n-4}{32n^3+4n^2}}{\sqrt{\frac{2n^3+3n-1}{{8n^3+n^2}}}+\frac{1}{2}}\right|\\ &<\left|\frac{\frac{-n^2+12n-4}{32n^3+4n^2}}{\frac{1}{2}}\right|\\ &=2\left|\frac{1-12\frac{1}{n}+4\frac{1}{n^2}}{32n+4}\right|\\ &<\frac{1}{16n+2} \end{align}$$

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

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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:

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

Answer 2

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