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

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#1
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.
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1 year ago
#2
(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"?
Last edited by ghostwalker; 1 year ago
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#3
(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
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1 year ago
#4
(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.
Last edited by ghostwalker; 1 year ago
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#5
(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
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#6
(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
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1 year ago
#7
(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?
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1 year ago
#8
(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.
Last edited by ghostwalker; 1 year ago
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#9
(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.
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1 year ago
#10
(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 as where the sgn function is 1 for h>0, and -1 for h <0

and this equals Clearly what's inside the modulus signs goes to zero as h goes to zero from either side, and so the whole thing does.
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#11
(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?
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1 year ago
#12
(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.
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#13
I got that:
|x|^n is always differentiable at x=0 for n>1 Because they converge to zero.
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1 year ago
#14
(Original post by Gjmvfhbk)
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
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1 year ago
#15
(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 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.
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