The pedantic function $\frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}$ I was shown the following "pedantic" function. 
$$
    f(x,y) := \frac{y \cdot \sin(x^5y^3+x^3)}{(x^4y^8+x^6+3y^2)\cos(x^2y)^2}
$$
The question is what happens as the function approaches origo. So in order for the limit to exists we need to check that the function is approaching the same limit for all directions. 
So along the coordinate axes we have
$$
  \lim_{x\to 0} f(x,0) = \lim_{y\to 0} f(y,0) =0
$$
Hmmm. Okay, lets test all straight lines through origo $y = k x$, some calculations again give
$$
\lim_{x \to 0} f(x,k x) = 0
$$ 
So it seems the limit exists, further studies show that
$$
\lim_{x \to 0} f(x,k x^n) = 0
$$ 
For all integers $n$, except $3$. Is this true?

Conjecture
  Let $f(x,y)$ be defined as above then
  $$
    \lim_{x\to 0} f(x,k x^n) = 0
$$
  for all $n \in\mathbb{N}$ except $n=3$. Here we have
  $$
   \lim_{x\to 0} f(x, k x^3) = \frac{k}{3k^2+1}
$$
  where $k \in \mathbb{R}$.

Is the conjecture true? I tested $n$ up to 2000. Also are
there other lines approaching origo that leads to a different answer than zero, except $y = x^3$? Can this be proven rigorously? 
 A: The conjecture is true, as you can see by finding the Taylor series for 
$$\frac{k x^3  \sin(x^3 + k^3 x^{14})}{x^6 + 3 k^2 x^6 + k^8 x^{28})\cos^2(kx^5)} = \frac{k}{1+3k^2}-\frac{kx^6}{6(1+3k^2)} + \frac{k^3x^{10}}{1+3k^2} + \dots$$
A: First, notice that 
$$f(x,kx^n) = \frac{kx^n \cdot \sin(x^5(kx^n)^3+x^3)}{(x^4(kx^n)^8+x^6+3(kx^n)^2)\cos(x^2(kx^n))^2} = \frac{kx^n\sin(x^3+k^3x^{n+5})}{(x^6+3k^2x^{2n}+k^8x^{8n+4})[\cos(kx^{n+2})]^2}$$
As $x\to 0$, lower powers of $x$ converge to $0$ faster than higher powers. So only the lowest order powers of $x$ actually matter in the limit (this is dual to the fact that as $x\to \infty$ only the highest powers matter).
Now $\sin(Z) = Z-\frac{Z^3}{3!}+\frac{Z^5}{5!}+\cdots$ so $\sin(x^3+k^3x^{n+5})=(x^3+k^3x^{n+5})+\cdots = x^3+O(x^4)$ 
If you're not familiar with "big-O" notation, $O(x^4)$ means that all of the rest of the terms are of order $x^4$ or greater. They go to $0$ faster than $x^3$, so they won't matter in the limit.
Likewise, $\cos(Z) = 1-\frac{Z^2}{2!}+\cdots$ so $\cos(kx^{n+2}) = 1-\frac{k^2x^{2n+4}}{2!}+\cdots = 1+O(x)$. 
Next, assuming that $n>3$, we have that $k^8x^{8n+4}$ and $3k^2x^{2n}$ are of higher order than $x^6$. So we get
$$f(x,kx^n) = \frac{kx^n(x^3+O(x^4))}{(x^6+O(x^7))(1+O(x))^2} = \frac{kx^{n+3}+O(x^{n+4})}{x^6+O(x^7)} \to 0$$
as $x \to 0$ since $x^{n+3}$ is a higher power than $x^6$ (since $x>3$).
So, yes, your conjecture does hold. :)
