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How is |x|^(5/4) differentiable at x=0?

I thought this was not differentiable at x=0 as the two sided limits don't agree, but the answers say it is differentiable. I'm confused.

Reply 1

ghostwalker

Original post by Gjmvfhbk

I thought this was not differentiable at x=0 as the two sided limits don't agree, but the answers say it is differentiable. I'm confused.

What's your justification for the statement "two sided limits don't agree"?

(edited 2 years ago)

Reply 2

Gjmvfhbk

Original post by ghostwalker

What's your justification for the statement "two sided limits don't agree"?

lim as h tends to zero of (f(x+h)-f(x))/h
Is |h|^5/4 ÷ h.
When h approaches zero from negative side f(h) is negative and it's positive when approached from positive side

Reply 3

ghostwalker

Original post by Gjmvfhbk

lim as h tends to zero of (f(x+h)-f(x))/h
Is |h|^5/4 ÷ h.
When h approaches zero from negative side f(h) is negative and it's positive when approached from positive side

In red, yes, it's ~~positve~~ negative for h<0, but what does it converge to as h goes to 0? Similarly for h>0

Edit: And I assume you're refering to the function in red, rather than f(h) in your final sentence.

Edit2: See strike.

(edited 2 years ago)

Reply 4

Gjmvfhbk

Original post by ghostwalker

In red, yes, it's positve for h<0, but what does it converge to as h goes to 0? Similarly for h>0

How is it positive for h<0?
|h|^(5/4) is positive and h is negative hence ((|h|^(5/4))/h is negative

Reply 5

Gjmvfhbk

Original post by ghostwalker

In red, yes, it's positve for h<0, but what does it converge to as h goes to 0? Similarly for h>0

Edit: And I assume you're refering to the function in red, rather than f(h) in your final sentence.

Original post by Gjmvfhbk

How is it positive for h<0?
|h|^(5/4) is positive and h is negative hence ((|h|^(5/4))/h is negative

Also |x| is not differentiable at x =0 for same reason. Because lim |h|/h is -1 for h<0 and 1 for h>0 hence two sided limits don't agree just like in the first example

Reply 6

ghostwalker

Original post by Gjmvfhbk

How is it positive for h<0?
|h|^(5/4) is positive and h is negative hence ((|h|^(5/4))/h is negative

Sorry, I meant negative. Your function in red - see my previous post - is negative for h<0, and positive for h>0.

BUT what, in each case, does that function converge to as h goes to 0?

Reply 7

ghostwalker

Original post by Gjmvfhbk

Also |x| is not differentiable at x =0 for same reason. Because lim |h|/h is -1 for h<0 and 1 for h>0 hence two sided limits don't agree just like in the first example

True, |x| is not differentiable at x=0, but that tells you nothing about whether |x|^(5/4) is differentiable or not at x=0.

(edited 2 years ago)

Reply 8

Gjmvfhbk

Original post by ghostwalker

True, |x| is not differentiable at x=0, but that tells you nothing about whether |x|^(5/4) is differentiable or not at x=0.

The answers says is converges to zero, although I don't know how.
Can you please tell me how, I am confused out of my mind.

Reply 9

ghostwalker

Original post by Gjmvfhbk

The answers says is converges to zero, although I don't know how.
Can you please tell me how, I am confused out of my mind.

We can rewrite

\dfrac{|h|^{5/4}}{h}

\operatorname{sgn(h)}\dfrac{|h|^{5/4}}{|h|}

where the sgn function is 1 for h>0, and -1 for h <0

and this equals

Unparseable latex formula:

\operatorname{sgn(h)}|h|^{1/4}}

Clearly what's inside the modulus signs goes to zero as h goes to zero from either side, and so the whole thing does.

Reply 10

Gjmvfhbk

Original post by Gjmvfhbk

The answers says is converges to zero, although I don't know how.
Can you please tell me how, I am confused out of my mind.

Also for any integer value 'n'
Would |x|^n be differentiable at x=0 Because they converge to 0?

Reply 11

ghostwalker

Original post by Gjmvfhbk

Also for any integer value 'n'
Would |x|^n be differentiable at x=0 Because they converge to 0?

Make sure you understand the working for your original example before you start generalising, and when you do, you should be able to answer that question yourself.

Reply 12

Gjmvfhbk

Original post by ghostwalker

We can rewrite

\dfrac{|h|^{5/4}}{h}

\operatorname{sgn(h)}\dfrac{|h|^{5/4}}{|h|}

where the sgn function is 1 for h>0, and -1 for h <0

and this equals

Unparseable latex formula:

\operatorname{sgn(h)}|h|^{1/4}}

Clearly what's inside the modulus signs goes to zero as h goes to zero from either side, and so the whole thing does.

Thanks for your help.
I got that:
|x|^n is always differentiable at x=0 for n>1 Because they converge to zero.

Reply 13

ghostwalker

Original post by Gjmvfhbk

Thanks for your help.
I got that:
|x|^n is always differentiable at x=0 for n>1 Because they converge to zero.

Yep. And n doesn't even have to be an integer. Any value > 1

Reply 14

RDKGames

Study Forum Helper

Original post by Gjmvfhbk

I thought this was not differentiable at x=0 as the two sided limits don't agree, but the answers say it is differentiable. I'm confused.

Differentiable at 0 if

\displaystyle \lim_{h\to 0_-} \dfrac{|h|^{5/4}}{h} = \lim_{h\to 0_+} \dfrac{|h|^{5/4}}{h}

The numerators have higher power of h than the denominators.

So it should be clear that both limits will be swayed to 0 due to numerators going to zero faster than denominators.

Hence the equality above holds, and there is differentiability.