Define a function $f$ by
$$
f(x,y)=\frac{x^3-2y^3}{x^2+2y^2}.
$$
I want to determine with proof whether or not $$\lim_{(x,y) \to (0,0)}f(x,y)$$ exists.
Here's what I've tried:
I want to find a possible numerical value for the limit so that I can potentially prove that a limit exists with the Squeeze Theorem.
To do so, I approach the point $(0,0)$ along lines of the form $y=mx$ so that the limit becomes
$$
\lim_{x\to 0}\frac{x^3-2m^3x^3}{x^2+2m^2x^2}=\lim_{x\to 0}\frac{x^3(1-2m^3)}{x^2(1+2m^2)}=\lim_{x\to 0}\frac{x(1-2m^3)}{1+2m^2}=0.
$$
Now that I have a potential limit of $L=0$, I want to "set up" the Squeeze Theorem by finding a $g(x,y)$ such that
$$
|f(x,y)-L|\le g(x,y).
$$
Here's my problem:
I'm failing to find a $g(x,y)$ that makes $\lim_{(x,y) \to (0,0)}g(x,y)$ easily solvable.
This suggests to me that the limit doesn't actually exist.
In this case, I want to show that the limit does not exist by approaching $(0,0)$ along smooth curves and obtaining a limit value that is not $L=0$, violating the unique property of limits.
How is my approach to this problem? I feel that I'm stuck in a loop of believing the limit exists and trying to find a $g(x,y)$ and believing the limit does not exist and trying to choose some curve at random to approach $(0,0)$ along.
Any suggestions? Thanks!
Best Answer
HINT
I would start with noticing that \begin{align*} x^{2} \leq x^{2} + 2y^{2} \Rightarrow \left|\frac{x^{2}}{x^{2} + 2y^{2}}\right|\leq 1 \Rightarrow \left|\frac{x^{3}}{x^{2} + 2y^{2}}\right| \leq |x| \end{align*} Similarly, the following relation holds: \begin{align*} 2y^{2} \leq x^{2} + 2y^{2} \Rightarrow \left|\frac{2y^{2}}{x^{2} + 2y^{2}}\right| \leq 1 \Rightarrow \left|\frac{2y^{3}}{x^{2} + 2y^{2}}\right| \leq |y|\end{align*}
Now you can apply the squeeze theorem to: \begin{align*} \left|\frac{x^{3} - 2y^{3}}{x^{2} + 2y^{2}}\right| \leq \left|\frac{x^{3}}{x^{2} + 2y^{2}}\right|+ \left|\frac{2y^{3}}{x^{2} + 2y^{2}}\right| \leq |x| + |y| \end{align*}
Can you take it from here?