# Quod Erat Demonstrandum

## 2013/04/24

### To f.4 M2 of my group

Filed under: NSS — johnmayhk @ 3:05 下午

In the recent sudden quiz, there is a so-called bonus question (actually it is a boring 1-mark question): evaluate

$\displaystyle \lim_{n\rightarrow \infty}(\sqrt[8]{n^2+1}-\sqrt[4]{n+1})$.

Here is a suggested solution

$\displaystyle \lim_{n\rightarrow \infty}(\sqrt[8]{n^2+1}-\sqrt[4]{n+1})$

$=\displaystyle \lim_{n\rightarrow \infty}\frac{\sqrt[4]{n^2+1}-\sqrt{n+1}}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}$

$=\displaystyle \lim_{n\rightarrow \infty}\frac{\sqrt{n^2+1}-(n+1)}{(\sqrt[8]{n^2+1}+\sqrt[4]{n+1})(\sqrt[4]{n^2+1}+\sqrt{n+1})}$

$=\displaystyle \lim_{n\rightarrow \infty}\frac{(n^2+1)-(n+1)^2}{(\sqrt[8]{n^2+1}+\sqrt[4]{n+1})(\sqrt[4]{n^2+1}+\sqrt{n+1})(\sqrt{n^2+1}+(n+1))}$

$=\displaystyle \lim_{n\rightarrow \infty}\frac{-2n}{(\sqrt[8]{n^2+1}+\sqrt[4]{n+1})(\sqrt[4]{n^2+1}+\sqrt{n+1})(\sqrt{n^2+1}+(n+1))}$

$=\displaystyle \lim_{n\rightarrow \infty}\frac{-2}{(\sqrt[8]{n^2+1}+\sqrt[4]{n+1})(\sqrt[4]{n^2+1}+\sqrt{n+1})(\sqrt{1+\frac{1}{n^2}}+(1+\frac{1}{n}))}$

$=0$

Or, you may think about that

$\sqrt[8]{n^2+1}-\sqrt[4]{n+1}$

$=\sqrt[8]{n^2+1}-\sqrt[8]{n^2}+\sqrt[8]{n^2}-\sqrt[4]{n+1}$

$=(\sqrt[8]{n^2+1}-\sqrt[8]{n^2})+(\sqrt[4]{n}-\sqrt[4]{n+1})$

and consider that both

$\sqrt[8]{n^2+1}-\sqrt[8]{n^2}$ and $\sqrt[4]{n}-\sqrt[4]{n+1}$

tend to zero as $n$ tends to positive infinity, done.

[OT]

@_@

「……的校本評核。」