# Quod Erat Demonstrandum

## 2008/03/01

### Something to say about F.7 Pure Mathematics (II) Mock Exam (part 2)

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:02 下午

There was a piece of information given in the “順帶二提" at the last post “Taylor’s polynomials@濟濟一堂", that is

$f(x) = 0$ when $x = 0$ and
$f(x) = e^{\frac{-1}{x^2}}$ when $x \ne 0$

then $f^{(n)}(0) = 0$ for any $n \in \mathbb{N}$

This gave me the idea of setting the following standard short question Q.2 in the mock paper.

$f(x) = 0$ when $x = 0$ and
$f(x) = e^{\frac{-1}{|x|}}$ when $x \ne 0$

Show that $f(x)$ is differentiable at $x = 0$.

May be we need to evaluate the following in the solution

$lim_{h \rightarrow 0^+}\frac{e^{-1/h}}{h}$.

By using l’ Hôpital’s rule directly, the situation ends with a ‘loop’ (just try).

Then I tried to prove $\lim_{h \rightarrow 0^+}\frac{1}{he^{1/h}} = 0$ by showing $\lim_{h \rightarrow 0^+}he^{\frac{1}{h}} = +\infty$. To do so, I try to write a lower bound of $e^x$, with just a bit expansion, I choose $e^x \ge 1 + x + \frac{x^2}{2}$, then $he^{\frac{1}{h}} \ge h(1 + \frac{1}{h} + \frac{1}{2h^2}) = h + 1 + \frac{1}{2h} \rightarrow +\infty$ as $h \rightarrow 0^+$. Afterwards, I think it is not a good method because we need the extra help “$e^x \ge 1 + x + \frac{x^2}{2}$“, then I tried to do it by l’ Hôpital’s rule merely, just use the substitution $u = \frac{1}{h}$, then the limit can be found easily, how amazing, isn’t it?

## 2 則迴響 »

1. 感謝你，John Sir。

的確，也許是我太過執著於所謂的「理性爭論」。其實再爭下去，別人也有自己的一套思想理論，自己也有一套自圓其說的法則。而且，即使你在辯論中勝了或是輸了，對自己也沒有好處。

的確，順服了主基督，看到的人和事，與「世俗」確實有點不同。自己經歷的感覺也很實在。也許是主耶穌的愛感動了我們，使我們成了基督的門徒。

也許是自己的自信使自己遠離神，甚麼哲學、科學的爭辯只是自己給自己的藉口。可能我對主的付主太少，信心也就不夠。

對。為你為我禱告。當人覺得自己有能力應付一切的時候，便會忘記神。

迴響 由 wonghon — 2008/03/02 @ 4:32 下午 | 回應

2. lim (h->0+) (1/h)/e^(1/h)

迴響 由 y.u. — 2008/03/21 @ 7:54 下午 | 回應