Need help showing this difference quotient is a derivative

[Eclair_de_XII]

Homework Statement

"Suppose $f:(a,b) \rightarrow ℝ$ is differentiable at $x\in (a,b)$. Prove that $lim_{h \rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}$ exists and equals $f'(x)$. Give an example of a function where this limit exists, but the function is not differentiable."

Homework Equations

Differentiability: Let $f:D\rightarrow ℝ$ with $x_0\in D'\cap D$ ($D'$ represents the set of $D$'s accumulation points). For each $t\in ℝ$ such that $x_0+t \in D$ and $t\neq 0$, define $Q(t)=\frac{f(x_0+t)-f(x_0)}{t}$. The function $f$ is said to be differentiable at $x_0$ iff $Q$ has a limit at zero.

The Attempt at a Solution

I know it's not really anything at all, but this is what I came up with for the first part:

$lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}=lim_{h\rightarrow 0} \frac{f((x-h)+2h)-f(x-h)}{2h}=f'(x-h)$

 

[fresh_42]

You have still a limit in $f\,'(x-h)$ which makes it a left derivative. Now can you do the same from the other side? Then you could use that differentiable means that both - left and right - have to be the same value.

 

[Eclair_de_XII]

Let's see...

$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2} \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}+\frac{1}{2} \lim_{h\rightarrow 0} \frac{f(x)-f(x-h)}{h}=\frac{1}{2}f'(x)+\frac{1}{2}f'(x-h)=f'(x)$

 

[fresh_42]

Yes, just without the last $h$.

 

[Eclair_de_XII]

Thanks. Now, I just need to figure out the second part of this problem, then. Just give me some time to think...

 

[fresh_42]

Thanks. Now, I just need to figure out the second part of this problem, then. Just give me some time to think...

We used: differentiable implies left and right derivatives are equal. There's an easy, very common function which violates this at one point.

 

[Eclair_de_XII]

Oh, you mean: $f(x)=|x|$?

 

[fresh_42]

Oh, you mean: $f(x)=|x|$?

You might need to change the slope of one leg to save the $h$ in the nominator.

 

[Eclair_de_XII]

Can I ask what you mean by "change the slope of one leg"?

 

[fresh_42]

Can I ask what you mean by "change the slope of one leg"?

If we simply use $f(x)=|x|$ then $\lim_{h \to 0}=\dfrac{f(x+h)-f(x-h)}{2h}=\lim_{h\to 0}\dfrac{h-h}{2h}$ at $x=0$. I think in this case, we have to formally show that this limit exists. It's easier to manipulate $f$ in a way, that the nominator is e.g. $h-\frac{1}{2}h$ and we can divide by $h$.

 

[Eclair_de_XII]

I think I got something...

Let $\lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}=\lim_{h\rightarrow 0} \frac{|x+h|-|x-h|}{2h}=L$.

Then $\lim_{h\rightarrow 0} \frac{|x+h|-|x-h|}{2h}\leq \lim_{h\rightarrow 0} |\frac{(x+h)-(x-h)}{2h}|=\lim_{h\rightarrow 0} |\frac{2h}{2h}|=1$.
And $-1=-\lim_{h\rightarrow 0} |\frac{(x+h)-(x-h)}{2h}| \leq \lim_{h\rightarrow 0} \frac{|x+h|-|x-h|}{2h}$.

So $-1\leq L \leq 1$, so $\lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}=L$ exists, but $f(x)=|x|$ has no derivative at $x=0$.

 

[fresh_42]

I don't see why $L$ should exist by your argumentation. You started by the assumption that it exists. That it is in $[-1,1]$ under this assumption, cannot be used to prove the assumption. At $x=0$ we can directly estimate $| \frac{\Delta f}{h} - 0| < \varepsilon$ for small $h$. Thus $\lim_{h \to -0} \Delta f = -1 \neq \lim_{h \to 0}\Delta f = 0 \neq \lim_{h \to +0}\Delta f=1\,.$

I still find it easier to see, if we defined $f(x)=|x|$ for $x \leq 0$ and $f(x)=5|x|$ for $x>0\,$, and get $\lim_{h \to 0} \Delta f = 2$ because the suspicious $h$ in the denominator doesn't occur; but both ways would be o.k.

 

[StoneTemplePython]

there is is a very special, but quite common class of functions that give you the second leg of your answer but do not vex yourself over this

and yes the absolute value is a member of this class