Proving that a limit does not exist

[noelo2014]

Homework Statement

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Homework Equations

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The Attempt at a Solution

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Basically I need to know if the way I've done this problem is OK. I approached the point (0,0) along path y=x², substituted 0 for x and got a value of 1. Fine

Then I approached along the Y axis, meaning x is always 0. Because x is always 0 and the numerator, the value of the function will be zero for all y.

Is this an OK argument?

Thanks

 

[haruspex]

Your argument is sound. (But why not just approach along y=x?)

 

[noelo2014]

Your argument is sound. (But why not just approach along y=x?)

Because there's an x-y on the bottom, therefore this will give a 0 on the denominator for all x

 

[haruspex]

Because there's an x-y on the bottom, therefore this will give a 0 on the denominator for all x

Right. So? Doesn't that prove there's no such limit?

 

[noelo2014]

Right. So? Doesn't that prove there's no such limit?

So are you saying if the limit is not continuous in a neighbourhood of (x0,y0) then that's sufficient proof that the limit does not exist?

 

[haruspex]

So are you saying if the limit is not continuous in a neighbourhood of (x0,y0) then that's sufficient proof that the limit does not exist?

I'm saying that if you have an infinite sequence of points converging to (x0, y0) such that the function is not defined at any of them then it cannot be continuous at (x0, y0).

 

[noelo2014]

I'm saying that if you have an infinite sequence of points converging to (x0, y0) such that the function is not defined at any of them then it cannot be continuous at (x0, y0).

Ok, I think that's what I meant. I have another question which I've been stuck on for hours: How do I find the derivative with respect to x of

x^sin(x)

I'm really stuck on this, been looking for a good explanation on how to do it. I've a calculus exam tomorrow

 

[Dick]

Ok, I think that's what I meant. I have another question which I've been stuck on for hours: How do I find the derivative with respect to x of

x^sin(x)

I'm really stuck on this, been looking for a good explanation on how to do it. I've a calculus exam tomorrow

$x=e^{log(x)}$. So $x^{sin(x)}=e^{log(x) sin(x)}$. Now use the chain rule on that. In the future, I'd open a new thread for unrelated questions.