Is a Function Differentiable if Its Symmetric Derivative Exists?

[Lily@pie]

Homework Statement

If a function satisfies g'(x) = lim_(h->0) {[g(x+h)-g(x-h)/2h}, must g be differentiable at x? Provide a proof or counter example

Homework Equations

From the formal definition of differentiation, I know that g'(x) = lim _(h->0) {[g(x+h)-g(x)]/h}

The Attempt at a Solution

I think that g must be differentiable near x but not at x.

As differentiability implies continuity, if g is not differentiable at x, the point g(x) is not defined at x. However, the lim_(h->0) {[g(x+h)-g(x-h)/2h} will still exist as we don't care about the limit at that particular point but near that point.

But I just couldn't find a counter example that represent this situation. Or, am I wrong?

Thanks

 

[micromass]

If a function satisfies g'(x) = lim_(h->0) {[g(x+h)-g(x-h)/2h}

I don't understand something. You say that g'(x) is that limit, but what is g'(x)? If it is the derivative, then there's something fishy, since then you assume that g'(x) exists...

But for your question: can you give me a function that is not differentiable at a point? Can you check the condition on that function?

 

[CompuChip]

Edit: micromass is right, I assumed that you meant: "If, for a function g, lim(h->0) {[g(x+h)-g(x-h)/2h} exists and g is differentiable, show that g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h}".

I think that g must be differentiable near x but not at x.

What does "near x" mean? I mean, can you express that with a limit?

As differentiability implies continuity, if g is not differentiable at x, the point g(x) is not defined at x.

That is not true. Differentiability implies continuity. So if g is differentiable at x, then g is continuous at x. The converse statement is: if g is not continuous, then it is also non-differentiable.
As a counter-example to your statement, take g(x) = |x| (the absolute value function). It is non-differentiable at x = 0, yet g(x) = 0.

The Attempt at a Solution

Try rewriting
lim(h->0) {[g(x+h)-g(x-h)]/2h}
as
lim(h->0) {(g(x+h)-g(x))-(g(x-h) - g(x))]/2h}

 

[Lily@pie]

I don't understand something. You say that g'(x) is that limit, but what is g'(x)? If it is the derivative, then there's something fishy, since then you assume that g'(x) exists...

But for your question: can you give me a function that is not differentiable at a point? Can you check the condition on that function?

g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h} is the condition given in the question. And if I draw the graph on it, lim(h->0) {[g(x+h)-g(x-h)/2h} is the same as lim (h->0) {[g(x+h)-g(x)]/h}; which is the formal definition of g'(x)

I couldn't find a counter example actually. But I couldn't find a way to prove too.

 

[micromass]

g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h} is the condition given in the question. And if I draw the graph on it, lim(h->0) {[g(x+h)-g(x-h)/2h} is the same as lim (h->0) {[g(x+h)-g(x)]/h}; which is the formal definition of g'(x)

I couldn't find a counter example actually. But I couldn't find a way to prove too.

Well, if that's the question then the question is ill-posed. You can't use g'(x) without knowing that g is differentiable... I have no idea what they're trying to say...

 

[Lily@pie]

Edit: micromass is right, I assumed that you meant: "If, for a function g, lim(h->0) {[g(x+h)-g(x-h)/2h} exists and g is differentiable, show that g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h}".


What does "near x" mean? I mean, can you express that with a limit?


That is not true. Differentiability implies continuity. So if g is differentiable at x, then g is continuous at x. The converse statement is: if g is not continuous, then it is also non-differentiable.
As a counter-example to your statement, take g(x) = |x| (the absolute value function). It is non-differentiable at x = 0, yet g(x) = 0.



Try rewriting
lim(h->0) {[g(x+h)-g(x-h)]/2h}
as
lim(h->0) {(g(x+h)-g(x))-(g(x-h) - g(x))]/2h}

For instance, the limit when approaching x a little bit from the left is the same as the limit when approaching x a little bit from the right must be equal for the limit to exist. But it is not necessary for the function to be defined at x.

The question want us to prove that g must be differentiable at x if g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h}. I assume they want us to prove the limit exist?

 

[Lily@pie]

Well, if that's the question then the question is ill-posed. You can't use g'(x) without knowing that g is differentiable... I have no idea what they're trying to say...

I'm confused ><

 

[Lily@pie]

Because the question have 2 parts.
The 1st part is assuming g is differentiable at x, prove g'(x) = lim(h->0) {[g(x+h)-g(x-h)/2h}. -- (1)
And I have proven that.

The 2nd part is for a function that satisfy (1) at some point x, must g be differentiable at x?

 

[Lily@pie]

Okie... I'm so sorry...

I asked my lecturer about it and he say the question is phrased wrongly...

It should be if lim(h->0) {[g(x+h)-g(x-h)/2h}, must g be differentiable at x?

 

[HallsofIvy]

Okie... I'm so sorry...

I asked my lecturer about it and he say the question is phrased wrongly...

It should be if lim(h->0) {[g(x+h)-g(x-h)/2h}, must g be differentiable at x?

You mean, I think, that if lim_{h\to 0}[g(x+h)- g(x- h)/2 exists, must g be differentiable at x?

Look at g(x)= |x|.

 

[Lily@pie]

You mean, I think, that if lim_{h\to 0}[g(x+h)- g(x- h)/2 exists, must g be differentiable at x?

Look at g(x)= |x|.

for g(x) = |x|

lim_(h->0) {[g(x+h)-g(x-h)/2h} = 1

but
g'(x) = lim _(h->0) {[g(x+h)-g(x)]/h} does not exist.

So, it is not necessary for g to be differentiable... Right?

Thanks so much.

Just curious, are there any reasons for this situation? Because I always thought
lim_(h->0) {[g(x+h)-g(x-h)/2h}= lim _(h->0) {[g(x+h)-g(x)]/h}

 

[Lily@pie]

Oh I'm so sorry. I haven't learn differentiation for abs values.

I recalculated but lim(h->0) {[g(x+h)-g(x-h)/2h} is also undefined when x=0...

My calculations:
lim(h->0) {[|x+h|-|x-h|]/2h}
When x=0
=lim(h->0) {[|h|-|-h|]/2h}
when |h| - |-h|, it is 0, but 2h is also 0.

What did i do wrong?

 

[HallsofIvy]

Yes,
\frac{0}{2h}= 0
as long as h is not 0. Therefore the limit is 0. It does not matter that you get "0/0" at h= 0. What happens at x= a is irrelevant to the limit \lim_{x\to a} f(x).

(In taking the limit of the difference quotient to find the derivative, the denominator always goes to 0!)

\lim_{h\to 0}\frac{|0+h|- |0- h|}{2h}= \lim_{h\to 0} \frac{0}{h}= 0

\lim_{h\to 0}\frac{|0+h|- |0|}{h}
does NOT exist because the two "onesided" limits are not the same:

\lim_{h\to 0^+}\frac{|h|}{h}= \lim_{h\to 0^+}\frac{h}{h}= \lim_{h\to 0^+} 1= 1
because, with h approaching 0 from above, h> 0 and |h|= h.

But
\lim_{h\to 0^-}\frac{|h|}{h}= \lim_{h\to 0^-}\frac{-h}{h}= \lim_{h\to 0^-} -1= -1
because, with h approaching 0 from below, h< 0 and |h|= -h.