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]

那早晨聽著手機傳來的新聞報導:「……新高中檢討後,將取消數學、體育、企業會計及財務管理三科……..」

@_@

「……的校本評核。」

但有家長高興地告訴子弟:「不用擔心了,數學將不會是主科!」這當然不涉及主流媒體誤導,純粹是家長不清楚何謂校本評核而已。

身為數學授課員,「看」到現今學生對數學的表現與態度,包括在顏冊或別的地方對數學和/或數學授課員的咒罵,希望取消數學科的學生(或家長)似乎佔大多數。

聞說「生活化」有助提升學習數學之興趣,今屆 DSE 數學考卷問「保安禁區」面積夠不夠大,也問公屋總樓面面積是否「供過於求」,使我有所聯想,不知我是否太政治敏感?

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