Find two different "ways" to show that the following limit doesn't exist $\lim_{(x,y)\to (0,0 )} \frac{x^3y}{x^6 + y^2}$.

Question

$$\lim_{(x,y)\to (0,0 )} \frac{x^3y}{x^6 + y^2}$$

I tried to approach it by the $y$-axis:

$$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0 \cdot y}{0+y^2}=0$$

by the $x$-axis:

$$\lim_ {(x)\to (0),(y=0)} =\lim_ {(x) \to (0)}=\frac{x^3 \cdot 0}{x^6+0^2}=0$$

by $y=x$

$$\lim_ {(x)\to (0),(y=x)} =\lim_ {(x) \to (0)}=\frac{x^3+x}{x^6+x^2}=\frac{x^4}{x^6+x^2}=0$$

Is there any other "way" which gives a different result from $0$? Are my calculations correct?

Answer 1

(Community Wiki so that the question registers an answer.)

As above, approaching the origin along the curve $y = x^3$ suggests a limit that approaches $1/2$.

Since it is shown in the OP that other approaches yield a limit that suggests $0$, the limit of the expression cannot exist: $0 \neq 1/2$.

Answer 2

An other good method here is also to try with sequences. For instance here if I take $$(x_n,y_n) = (\frac{1}{n},\frac{1}{n^3})$$ which converges to 0 then $$\frac{x_n^3y_n}{x_n^6+y_n^2} = 1/2$$ But if I take $$(\frac{1}{n},\frac{1}{n}))$$ we find as a limit 0. The limits are different for two sequences that converge to 0 so the limit doesn't exist.

I specifically chose this sequence because you can see that the power of x in all its occurrence is 3 times the power of y in all its occurrences and since its a quotient it works out well. I don't know if that's well explained though, tell me if not.