1. ## Limits

find this
the picture

2. Originally Posted by rqeeb
find this

the picture
evaluate it along the path $y=mx^3$.

3. Originally Posted by Ted
evaluate it along the path $y=mx^3$.
it will be (m^3)/(1+m^4)

but if we use y=x it will be zero

which one is the correct

4. Originally Posted by rqeeb
it will be (m^3)/(1+m^4)

but if we use y=x it will be zero

which one is the correct
No need to use $y=x$; since the value of the limit along the path $y=mx^3$ depends on m, then the limit has different values for different values of $m$.
Hence, the limit does not exist.

5. Originally Posted by Ted
No need to use $y=x$; since the value of the limit along the path $y=mx^3$ depends on m, then the limit has different values for different values of $m$.
Hence, the limit does not exist.
thanks

but if we use polar coordinates it will be zero.... why

6. Originally Posted by rqeeb
thanks

but if we use polar coordinates it will be zero.... why
If you use polar coordinates, you will face:

$\lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12 }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{r^2sin^3(\theta)cos^3(\theta)}{r^8cos^{12}(\ theta)+sin^4(\theta)}$

What did you do for it ?

7. Originally Posted by Ted
If you use polar coordinates, you will face:

$\lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12 }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{r^2sin^3(\theta)cos^3(\theta)}{r^8cos^{12}(\ theta)+sin^4(\theta)}$

What did you do for it ?
this

$\lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12 }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{0}{0+sin^4(\theta)}=0$

8. Originally Posted by rqeeb
this

$\lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12 }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{0}{0+sin^4(\theta)}=0$
Nope. The problem here is that $sin^4(\theta)$ can be zero.
and this will make the limit $=\frac{0}{0}$ which is indeterminate.

9. Originally Posted by Ted
Nope. The problem here is that $sin^4(\theta)$ can be zero.
and this will make the limit $=\frac{0}{0}$ which is indeterminate.
Is that right!!
because our teacher told us that we deal with any term has theta as a constant.

10. Originally Posted by rqeeb
Is that right!!
because our teacher told us that we deal with any term has theta as a constant.
and the constant could be zero.

11. It could be
Thanks 4 everything