The question in place would have me figure out $$\lim_{(x,y)\to(0,0)}\frac{y^4 - 2x^2}{y^4 + x^2}$$ What I have tried already:
I made $y = mx$, and I decided since m is an fixed constant, of which I have
decided to make 0. So, $y = 0x$, therefore $y = 0$.
So, I plug $y = mx$ into the function, and I get $(mx^4 - 2x^2)/(mx^4 + x^2)$, I then factored out $x^2$ from the top and the bottom, leaving me with $(mx^2- 2)/(mx^2 +1)$ and since $m = 0$, and that means our $x$ is automatically 0 as well
for $y = 0$. So, plugging that in, we are left with -2/1 of which the limit is $-2$.
Hey, that's right! (According to the book.)
Now, here is where my problem comes up, I do the same thing for $x = my , m = 0, x = 0$. So, that gives us $(y^4 -2my)/(y^4 +2(my)^2)$. Factor out $y^2$ because we
can, and we get $(y^2 -2m^2) / ( y^2 + m^2)$ and now I'm at an impasse.
You can't have 0 in the denominator, and I have a zero in the denominator.
I'm honestly not sure how to deal with this, please assist senpai. Thank you.
You don't have a zero in the denominator, but $m=0$ and the expression reduces to $1$. Since the limit along $y=0$ was previously shown to be $-2$, the limit does not exist.