How to solve the limit of a multivariable function I have the following limit, when I try to approach it on the x-axis the result is 0, and when I try to approach it on the y-axis the result is 0, I can therefore not say if it exists or not.

          lim                 (x-4)3  (y-1)5 
(x, y) → (4, 1)     __________
                             (x-4)8 + (y-1)8 
switching to polar cordinates: 
x = r cost + 4
y = r cost + 1

          lim                 ((r cost + 4) - 4)3  ((r sin t + 1)-1)5 
       r → 0            ________________________
                             ((r cost + 4) - 4)8 + ((r sin t + 1) - 1)8 

          lim                 (r cost)3  (r sin t)5 
       r → 0            ______________
                             (r cost)8 + (r sin t)8 

          lim                 r3 cos3t  r5 sin 5t 
       r → 0            ______________
                             (r8 cos8t) + (r8 sin8t) 

          lim                 r8 cos3t  sin 5t 
       r → 0            ______________
                              r8((cos8t) + (sin8t)) 

          lim                 cos3t   sin 5t 
       r → 0            ______________
                              ((cos8t) + (sin8t)) 
In the last part, I no longer have r in my expression left. Does this mean that the limit does not exist?
 A: As mentioned in the post we switched the limit to a limit with polar coordinates using $$x=r\cos(\theta)+4 \quad \text{and} \quad y=r\sin(\theta)+1$$
then as $r\to 0$ we have $x\to 4$ and $y \to 1$.
This new limit is then
\begin{align*}\lim_{(x,y)\to (4,1)}\frac{(x-4)^{3}(y-1)^{5}}{(x-4)^{8}+(y-1)^{8}} &= \lim_{r\to0}\frac{r^{8}\cos^{3}(\theta)\sin^{5}(\theta)}{r^{8}\cos^{8}(\theta)+r^{8}\sin^{8}(\theta)}\\
&=\lim_{r\to0}\frac{\cos^{3}(\theta)\sin^{5}(\theta)}{\cos^{8}(\theta)+\sin^{8}(\theta)}\\
&=\frac{\cos^{3}(\theta)\sin^{5}(\theta)}{\cos^{8}(\theta)+\sin^{8}(\theta)}.
\end{align*}
Here we can see that we will get the limit to be $0$ if we approach either along the $y- axis$ or $x-axis$ (that is when $\theta$ is $\frac{\pi}{2}, \frac{3\pi}{2},0 \ \text{or} \ \pi$). However, pick any theta such that both $\cos(\theta)$ and $\sin(\theta)$ are not $0$ and you will get a nonzero limit. Therefore the limit does not exist.
Note: This is the answer to the question before it was edited to say something else.
A: By $u=x-4\to 0$ and $v=y-1\to 0$ we have
$$\lim_{(x,y)\to (4,1)}\frac{(x-4)^{3}(y-1)^{5}}{(x-4)^{8}+(y-1)^{8}}=\lim_{(u,v)\to (0,0)}\frac{u^{3}v^{5}}{u^{8}+v^{8}}$$
and by $v=u\to 0$
$$\frac{u^{3}v^{5}}{u^{8}+v^{8}}=\frac{u^{3}u^{5}}{u^{8}+u^{8}}=\frac{u^{8}}{2u^{8}}=\frac12 \to \frac12$$
while by $v=-u\to 0$
$$\frac{u^{3}v^{5}}{u^{8}+v^{8}}=\frac{-u^{3}u^{5}}{u^{8}+u^{8}}=-\frac{u^{8}}{2u^{8}}=-\frac12 \to -\frac12$$
therefore the limit doesn't exist.
