MHB Finding Limit of Multi Variable Function Using Squeeze Law

[Barioth]

Let $$f(x,y)= \frac{x^3+y^3}{x^2-y} $$if $$x^2 $$is not equal to y
and 0 otherwise.

I need to find the limit when (x,y)->(0,0)

I'm sure the limit exist and is equal to 0, I was unable to conclude switching to polar coordinate so I wana use "the squeeze law"

I only have to evaluate $$f(x,y)= \frac{x^3}{x^2-y} $$

Here is what I get:

$$0 \leq |f(x,y)|= \frac{x^2|x|}{|x^2-y|}$$ now if $$|x^2-y|\geq 1$$ I can conclude that f(x,y)->0 as (x,y)->0

Sadly I don't know how the handle the case where $$|x^2-y|<1$$

any hint?

Thanks

 

[Fermat1]

Let $$f(x,y)= \frac{x³+y³}{x²-y} $$if $$x² $$is not equal to y
and 0 otherwise.

I need to find the limit when (x,y)->(0,0)

I'm sure the limit exist and is equal to 0, I was unable to conclude switching to polar coordinate so I wana use "the squeeze law"

I only have to evaluate $$f(x,y)= \frac{x³}{x²-y} $$

Here is what I get:

$$0 \leq |f(x,y)|= \frac{x²|x|}{|x^2-y|}$$ now if $$|x²-y|\geq 1$$ I can conclude that f(x,y)->0 as (x,y)->0

Sadly I don't know how the handle the case where $$|x²-y|<1$$

any hint?

Thanks

Setting $x=r\cos{\alpha}$, $y=r\sin{\alpha}$, then the function reads $\frac{r^2(\cos^3({\alpha})+ \sin^3({\alpha}))}{r\cos^2({\alpha})-\sin({\alpha})}$ which is a continuous function of $r$ so we may plug in $r=0$ to get the limit, which is zero.

 

[chisigma]

Let $$f(x,y)= \frac{x³+y³}{x²-y} $$if $$x² $$is not equal to y
and 0 otherwise.

I need to find the limit when (x,y)->(0,0)

I'm sure the limit exist and is equal to 0, I was unable to conclude switching to polar coordinate so I wana use "the squeeze law"

I only have to evaluate $$f(x,y)= \frac{x³}{x²-y} $$

Here is what I get:

$$0 \leq |f(x,y)|= \frac{x²|x|}{|x^2-y|}$$ now if $$|x²-y|\geq 1$$ I can conclude that f(x,y)->0 as (x,y)->0

Sadly I don't know how the handle the case where $$|x²-y|<1$$

any hint?

Thanks

Unfortunately one of the possible way to arrive at [0,0] is along the line $\displaystyle y= x^{2}$ and that means that in any case the limit doesn't exist...

Kind regards$\chi$ $\sigma$

 

[Evgeny.Makarov]

Let $$f(x,y)= \frac{x³+y³}{x²-y} $$if $$x² $$is not equal to y
and 0 otherwise.

I need to find the limit when (x,y)->(0,0)

I'm sure the limit exist and is equal to 0

But $x^2$ in the denominator can be arbitrarily close to $y$ so that $x^2-y$ can be much smaller than $x^3+y^3$. For example, let $x=1/n$ and $y=1/n^2-1/n^4=(n^2-1)/n^4$. Then
\[
\frac{x^3+y^3}{x^2-y}=n^4\left(\frac{1}{n^3}+\frac{(n^2-1)^3}{n^{12}}\right)
=n+\frac{(n^2-1)^3}{n^8}\to\infty
\]
as $n\to\infty$.

P.S. Use $x^2$ in LaTeX instead of the Unicode symbol for square. Compare $x^2$ and $x²$.

 

[Fermat1]

But $x^2$ in the denominator can be arbitrarily close to $y$ so that $x^2-y$ can be much smaller than $x^3+y^3$. For example, let $x=1/n$ and $y=1/n^2-1/n^4=(n^2-1)/n^4$. Then
\[
\frac{x^3+y^3}{x^2-y}=n^4\left(\frac{1}{n^3}+\frac{(n^2-1)^3}{n^{12}}\right)
=n+\frac{(n^2-1)^3}{n^8}\to\infty
\]
as $n\to\infty$.

P.S. Use $x^2$ in LaTeX instead of the Unicode symbol for square. Compare $x^2$ and $x²$.

So where does my method fail? Thanks

 

[Opalg]

You can also approach the origin along the curve $y = x^2(1-x)$, which gives the limit $1$.

[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]

So where does my method fail? Thanks

[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]It is possible for a function to have radial limits in all directions (at some point), but not to have a limit at that point.

 

[Fermat1]

You can also approach the origin along the curve $y = x^2(1-x)$, which gives the limit $1$.

[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]
[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]It is possible for a function to have radial limits in all directions (at some point), but not to have a limit at that point.

I've only looked up to how to do this but this link How to find this limit?
suggest that provided the limit does not depend on ${\alpha}$, then the method is valid

 

[Opalg]

I've only looked up to how to do this but this link How to find this limit?
suggest that provided the limit does not depend on ${\alpha}$, then the method is valid

Just goes to show you shouldn't believe everything on the internet (Shake) (or indeed in a textbook – they can sometimes make mistakes too).

The radial limit can be the same from all directions, yet the limit may not exist. The reason (as chisigma pointed out in comment #3) is that you can approach the limit point along curves as well as along straight lines, and thereby get different limits. The problem in this thread is an illustration of that. The limit along every line through the origin is $0$, but if you approach the origin along a curve sufficiently close to the parabola $y=x^2$ then you can get a different limit.

 

[ThePerfectHacker]

Consider the following function,
$$ f(x,y) = \frac{xy^3}{x^2 + y^6} $$

Along the path $y=mx$ we get,
$$ f(x,mx) = \frac{x\cdot m^3x^3}{x^2 + m^6 x^6} = \frac{x^2m^3}{1 + x^4m^6} \to 0 \text{ as }x\to 0 $$

Thus, it has the same limit radially. However, along the path $x=y^3$ we get,
$$ f(y^3,y) = \frac{y^3 \cdot y^3}{y^6 + y^6} = \frac{1}{2} $$
This shows that $f(x,y)$ does not posses a limit at $(0,0)$.

This is just another example of when radial limits are insufficient.

 

[Fermat1]

I think I realize my error now. In simplifying the function, I assumed $r$ is non-zero, and then plugged in $r=0$ to obtain the limit.