Quod Erat Demonstrandum


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。





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

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

    迴響 由 y.u. — 2008/03/21 @ 7:54 下午 | 回覆

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