Which still has a strange horizontal part near the zero, which does not disappear when I zoom in.
Did you mean 'vertical' ?
That alone would be the proof you want because the limit wouldn't exist at x = 0.
Arhh! seen your graph. That horizontal bit is fake. Most plotters work by evaluating as x is gradually incremented and then joining the resulting points. So what you are seeing is the effect of joining two points across the gap where x = 0. Think about what the limit actually does and you'll see it tends to infinity (-infinity) as x tends to zero.
Bob
]]>Question #2.
Yes, I did a mistake in the final step of calculating the limits. Sorry about that. It should be:
Question #3.
Actually, I do not see why did you say that domain of the function is just positive numbers? I believe the domain is full R.
After all, we can do :
I am not sure how that now applies on y=x^(2/3) .
But here is another example: x^(5/3). It is differentiable around 0 and the derivative at 0 is 0.
]]>It depends on your definition of diffentiation. Some versions require the left limit to equal the right limit and call your example 'semi-diffentiable'. That's why I included the infinite gradient test as well.
http://en.wikipedia.org/wiki/Semi-differentiability
(Actually I agree with you .... but I have met the alternative definition.)
Bob
]]>That is not always the case. Look at the graph of f(x)=x^(3/2). The function is defined for x>0 only, but f'(0) exists and is equal to 0.
]]>It is simpler than that. Take a look at the graph. (i) Values left of x = 0 don't exist (ii) the gradient tends to infinity as x approaches 0 from the right.
And having the two limits (left of point and right of point) is not a test either. Consider the graph
y = -3 x < 0
y = + 3 x > 0
Approaching 0 from either side, the gradients are the same. But the function is undefined and discontinuous at x = 0.
Bob
]]>So my answer is:
Now lets find left and right limits separately:
From this we can see that left and right limits at point 0 are equal by an absolute value, but have opposite signs; therefore, the
does not exists.But according to the rules of differentiation
or in this case:My problem is: how
correlate with ?