# Quod Erat Demonstrandum

## 2009/04/17

### 可導性

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:23 下午
Tags: , ,

1. Put $x = 0$ into the expression of $f'(x)$ and it is undefined, then $f(x)$ is not differentiable at $x = 0$. True?
2. If $\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x)$, then $f(x)$ is differentiable at $x = 0$. True?
3. If $\lim_{x \rightarrow 0^-}f'(x)$ is finite, then the value of $\lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h}$ is also finite. True?

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Question 1

$x = 0$ 入表達式 $f'(x)$，而「計不到數」（undefined），並不一定代表”$f(x)$ is not differentiable at $x = 0$”.

$f(x) = x^2\sin\frac{1}{x}$　for　$x \ne 0$　and　$f(0) = 0$.

$f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x}$ – – – – – – (*)

$f(x)$ is differentiable at $x = 0$”的意思是「極限 $\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}$ 存在」。

$f(x) = x^2\sin\frac{1}{x}$　for　$x \ne 0$　and　$f(0) = 0$.

$\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}$

$f(h) = h^2\sin\frac{1}{h}$, $f(0) = 0$，故

$\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}$
$= \lim_{h \rightarrow 0}\frac{1}{h}(h^2\sin\frac{1}{h} - 0)$
$= \lim_{h \rightarrow 0}h\sin\frac{1}{h}$

$\sin\frac{1}{h}$ 有界（bounded），且 $\lim_{h \rightarrow 0}h = 0$，故 $\lim_{h \rightarrow 0}h\sin\frac{1}{h} = 0$，即

$\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h} = 0$

$f(x)$ is differentiable at $x = 0$

Question 2Question 3 可以一併考慮，隨便舉例

$f(x) = x + 1$　for　$x \ge 0$　and
$f(x) = \sin x - 1$　for　$x < 0$

$f'(x) = 1$　for　$x > 0$　and
$f'(x) = \cos x$　for　$x < 0$

$\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^-}\cos x = \cos 0 = 1$
$\lim_{x \rightarrow 0^+}f'(x) = \lim_{x \rightarrow 0^+}1 = 1$

$\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x)$

$\lim_{x \rightarrow 0^-}f(x) = \lim_{x \rightarrow 0^-}\sin x - 1 = \sin 0 - 1 = -1$
$\lim_{x \rightarrow 0^+}f(x) = \lim_{x \rightarrow 0^+}(x + 1) = 0 + 1 = 1$

$\lim_{x \rightarrow 0^-}f(x) \ne \lim_{x \rightarrow 0^+}f(x)$

$\lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h}$
$= \lim_{h \rightarrow 0^-}\frac{(\sin h - 1) - 1}{h}$
$= \lim_{h \rightarrow 0^-}\frac{\sin h - 2}{h}$
$= +\infty$

$\lim_{x \rightarrow 0^-}f'(x) = 1$ 　（finite）但
$\lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h} = +\infty$　（infinite）

$\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x)$

1. Let

$f(x) = x + 1$　for　$x \ge 0$　and
$f(x) = \sin x + a$　for　$x < 0$

Suppose $f(x)$ is differentiable at $x = 0$, evaluate $a$.

2. Let

$f(x) = x + 1$　for　$x \ge 0$　and
$f(x) = b\sin x$　for　$x < 0$

Will $f(x)$ be differentiable at $x = 0$ for some real number $b$ ?

3. Let

$f(x) = R(x)$　for　$x \ge 0$　and
$f(x) = L(x)$　for　$x < 0$

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

Is is true to write ”$R'(0) = L'(0)$”?

## 4 則迴響 »

1. sir, here is my ans:

1. a=+1
2. differentiable when b=+1
3. T

but i actually don’t know what exactly what f'(x) is
what is the different in meaning between
lim f'(x)
x->0-
and
lim (f(h)-f(0))/h
h->o-
and also
lim f(x)
x->0-

i don’t know which one to consider when doing calculation

迴響 由 harrison — 2009/04/17 @ 8:16 下午 | 回應

2. Harrison,

The answer to Q.2 is FALSE.

$f(x)$ is NOT differentiable for ANY real number $b$.

Just think about the graph of $f(x)$, it is ALWAYS discontinuous at $x = 0$ (even for $b = 1$). (Or prove it by simple algebraic computation)

What is $f'(x)$?

It is a limit.

By definition,

If the limit $\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$ EXISTS, we define it as

$f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$.

What is $\lim_{x \rightarrow 0}f'(x)$?

It is NOT $f'(0)$ in general.

By the definition above,

$f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$.

hence

$\lim_{x \rightarrow 0}f'(x)$
$= \lim_{x \rightarrow 0}\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$

You see, it is NOT $f'(0)$ in general, because, by the definition, $f'(0)$ is simply:

$f'(0) = \lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}$

SOMETIMES, $f'(0)$ is really equal to $\lim_{x \rightarrow 0}f'(x)$, as for example, when the function $f'(x)$ is continuous at $x = 0$.

Also, can you prove that the answer to Q.3 is TRUE?

迴響 由 johnmayhk — 2009/04/18 @ 6:12 下午 | 回應

3. For Question 1, I used to teach my students in the same way as yours before at the beginning, but later I found that the answer of Q1 is true, and I also got a very simple proof.

Let me first point out the problem of your counter-example.
If f(x) = x^2 sin 1/x if x is not 0 and f(0) = 0,
then what is f'(x)?
f'(x) = (the expression you have) if x is not 0 and f'(0) = 0
Therefore, when we put x = 0 , f'(0) = 0 which is defined.
[ What you showed in your counter-example is in fact,
lim {x -> 0} f'(x) . ]

Back to the proof of the statement.
Assume f(x) is differentiable at x = 0.
Then f'(0) exists (defined), which leads to a contradiction.
Therefore f'(x) is not differentiable at x = 0.

迴響 由 Cheng Wing Kuen — 2009/04/21 @ 8:09 上午 | 回應

4. Thank you Cheng Wing Kuen.

是我的表達出了問題。

在 Question 1 中，

Put $x = 0$ into the expression of $f'(x)$

當中我希望說的，不是 $f'(0)$

當中的 expression，其實指，把 $f(x)$ 進行求導後得出的 expression。

如上例

$f(x) = x^2\sin\frac{1}{x}$

利用求導法，得出的 expression 就是

$f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x}$

Question 1 想說的，就是 put $x = 0$ into the expression above.

它是 undefined 的。

它不是 $f'(0)$

我只是想指出，同學往往誤以為

「把 $f(x)$ D 完，再代個零入去，計唔到，就即係函數係 x = 0 個位 D 唔到」。

是我的表達出問題，”expression of $f'(x)$”確實可以指

$f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x}$　when　$x \ne 0$　and　$f'(x) = 0$　when　$x = 0$

那麼，當”Put $x = 0$ into the expression of $f'(x)$” 就被理解為 $f'(0)$ 時，

誠如 Cheng Wing Kuen 所言，

Question 1 is trivially true.

We can prove it simply by contrapositivity.

$f(x)$ is differentiable at $x = 0$　$\Rightarrow f'(0)$ exists"

is equivalent to

$f'(0)$ does not exist　$\Rightarrow f(x)$ is not differentiable at $x = 0$"

P.S.

Oh, Mr. cheng, are you the writer of “永權網頁"? I love the website, it is very useful.

迴響 由 johnmayhk — 2009/04/21 @ 2:46 下午 | 回應