Some quick questions (1 Viewer)

[asianese]

Want to get these out before sleepin:

Find the limit, or show it doesn't exist:



The usual tricks like factorising/cancelling, going polar, x=y, x=0, y=0 don't seem to work? Probs blanking out. I know it doesn't exist but I can't find a path which isn't 0.

Next:









I get the partial deriv is f'(x) but i definitely did something wrong

 

[seanieg89]

Want to get these out before sleepin:

Find the limit, or show it doesn't exist:



The usual tricks like factorising/cancelling, going polar, x=y, x=0, y=0 don't seem to work? Probs blanking out. I know it doesn't exist but I can't find a path which isn't 0.

Next:









I get the partial deriv is f'(x) but i definitely did something wrong

1. Looking at the line y=x is enough.

2. Use the definition of partial derivatives and Taylor's theorem in the form f(x+h)=f(x)+hf'(x)+h^2f''(x)/2+O(h^3).

 

[braintic]

Find the limit, or show it doesn't exist:



The usual tricks like factorising/cancelling, going polar, x=y, x=0, y=0 don't seem to work? Probs blanking out. I know it doesn't exist but I can't find a path which isn't 0.

The limit definitely depends on the path.
The expression can be shown to equal x+y^2/(x-y).
As the first term is approaching zero, we need only look at the second term to determine the limit.
If we follow the curve y=2x, the second term simplifies to -3x, which approaches 0.
If we follow the curve y=x^2, the second term simplifies to (x^3)/(1-x), which approaches negative infinity.
If we follow the curve y=sqrt(x), the second term simplifies to x/(x-sqrtx) which approaches 1.
By choosing other paths, I'm sure an infinite number of limits can be arrived at.

 

[seanieg89]

The limit definitely depends on the path.
The expression can be shown to equal x+y^2/(x-y).
As the first term is approaching zero, we need only look at the second term to determine the limit.
If we follow the curve y=2x, the second term simplifies to -3x, which approaches 0.
If we follow the curve y=x^2, the second term simplifies to (x^3)/(1-x), which approaches negative infinity.
If we follow the curve y=sqrt(x), the second term simplifies to x/(x-sqrtx) which approaches 1.
By choosing other paths, I'm sure an infinite number of limits can be arrived at.

along y=2x the second term simplifies to -4x, which does indeed tend to 0.

x^3/(1-x) also tends to 0, not negative infinity.

x/(x-sqrt(x)) also tends to 0, not 1.

 

[braintic]

Oops, I was taking limits to infinity.

 

[seanieg89]

If you really want to explicitly find such a curve which gives you a finite nonzero limit the key is that the singular behaviour is on the line y=x, so something like y=x^2+x will do.

No point on such a singular line can ever be a point of continuity though.