Limit question: "Lemma: If g(x) is differentiable, then L{g(x+h)} = g(x)."

[Steven G]

Please let L stand for limit as h goes to 0.

Lemma: If g(x) is differentiable, then L{g(x+h)} = g(x).

Proof: g(x+h) = [g(x+h) - g(x)] + g(x) = [g(x+h) - g(x)]h/h + g(x)

Then L g(x+h) = L{[g(x+h) - g(x)] + g(x)} = L {[g(x+h) - g(x)]h/h + g(x)} = L {[g(x+h) - g(x)]/h}* L(h) + L g(x) = g'(x)*0 + g(x) = g(x).

I have been using this lemma for a long while (to prove the product rule). Today I was wondering what if we removed the condition that g(x) is differentiable.

I was wondering if someone can provide me with an example of g(x) such that L g(x+h) \(\displaystyle \neq\) g(x) or a proof of the lemma with the condition that g(x) is not differentiable (I do not think this is possible).

Thanks!!

EDIT: The proof above fails if g(x) is not differentiable since L {[g(x+h) - g(x)]/h} * 0 = undefined * 0 \(\displaystyle \neq\) 0

 

[Steven G]

Please let L stand for limit as h goes to 0.

Lemma: If g(x) is differentiable, then L{g(x+h)} = g(x).

Proof: g(x+h) = [g(x+h) - g(x)] + g(x) = [g(x+h) - g(x)]h/h + g(x)

Then L g(x+h) = L{[g(x+h) - g(x)] + g(x)} = L {[g(x+h) - g(x)]h/h + g(x)} = L {[g(x+h) - g(x)]/h}* L(h) + L g(x) = g'(x)*0 + g(x) = g(x).

I have been using this lemma for a long while (to prove the product rule). Today I was wondering what if we removed the condition that g(x) is differentiable.

I was wondering if someone can provide me with an example of g(x) such that L g(x+h) \(\displaystyle \neq\) g(x) or a proof of the lemma with the condition that g(x) is not differentiable (I do not think this is possible).

Thanks!!

EDIT: The proof above fails if g(x) is not differentiable since L {[g(x+h) - g(x)]/h} * 0 = undefined * 0 \(\displaystyle \neq\) 0

So I have been thinking about what makes a function not have a derivative even at just a point. My answer is that the function has a corner point or is discontinuous at a point (either removable or non removable). If it has a corner point I can not see why the theorem will not hold. If it has a discontinuity at a point then it will have a left and hand limit. But what if the function is not differentiable anywhere because of corner points or if the function is discontinuous everywhere?

 

[Steven G]

So I have been thinking about what makes a function not have a derivative even at just a point. My answer is that the function has a corner point or is discontinuous at a point (either removable or non removable). If it has a corner point I can not see why the theorem will not hold. If it has a discontinuity at a point then it will have a left and hand limit. But what if the function is not differentiable anywhere because of corner points of if the function is discontinuous everywhere?

Ahha! The piecewise function g(x) = 1 if x is irrational and = 0 if x is rational will be my counter example.

I must confessed that the reason I did this on my own was because of the forums policy that you must show work. As we all know it absolutely works!