Limit of two variables function

Question

I think that $$\lim_{(x,y)\to(0,0)} \left(\frac{3}{2y}-\frac yx\right)^2+\left(\frac 1y-\frac yx\right)^2=+\infty$$

But I can’t find a way to prove it. I first tried expanding the squares to find a reduction, but it gave me nothing. I tried polar coordinates, but I’m also stuck this way. I tried to write some minorization using $\dfrac 1{(xy)^2}\geqslant\dfrac 4{(x^2+y^2)^2}$. Any idea?

Answer 1

Clearly, the limit cannot be negative, since it is the sum of the squares of two functions of $x$ and $y$. Try the following.

Suppose that $$\lim_{(x,y)\to (0,0)} \bigg(\frac{3}{2y}-\frac{y}{x}\bigg)^2$$ does not diverge to $\infty$. Then the limit $$\lim_{(x,y)\to (0,0)} \bigg(\frac{3}{2y}-\frac{y}{x}\bigg)\tag{1}$$ does not diverge to $\pm\infty$. However, since $1/2y$ must diverge to $\pm\infty$ as $y\to 0$, and since $$\frac{3}{2y}-\frac{y}{x}-\frac{1}{2y}=\frac{1}{y}-\frac{y}{x}$$ it follows that if the limit $(1)$ doesn't diverge to $\pm\infty$, then the following limit will: $$\lim_{(x,y)\to (0,0)} \bigg(\frac{1}{y}-\frac{y}{x}\bigg)\tag{2}$$ Thus, either the limit $(1)$ or the limit $(2)$ or both diverge to positive or negative infinite values, meaning that the sum of their squares $$\lim_{(x,y)\to (0,0)} \bigg(\frac{3}{2y}-\frac{y}{x}\bigg)^2+\bigg(\frac{1}{y}-\frac{y}{x}\bigg)^2$$ diverges to $\infty$.

Answer 2

You can rewrite the first expression as:

$\Big( \frac{3}{2y}-\frac{y}{x} \Big)^2=\Big( \frac{1}{2y}+\frac{1}{y}-\frac{y}{x} \Big)^2$

Hence:

$\Big( \frac{1}{2y}+\frac{1}{y}-\frac{y}{x} \Big)^2=\frac{1}{4y^2}+\frac{2}{2y}\Big(\frac{1}{y}-\frac{y}{x} \Big)+\Big( \frac{1}{y}-\frac{y}{x} \Big)^2-\Big( \frac{1}{y}-\frac{y}{x} \Big)^2=$

$=\frac{1}{4y^2}+\Big( \frac{1}{y^2}-\frac{1}{x} \Big)=\frac{1}{4y^2}+\frac{x-y^2}{xy^2}=\frac{x+4x-4y^2}{4xy^2}$

This is an expression easier to work with $r\rightarrow 0$.

Answer 3

I think I got it using @Frpzzd’s idea. Let $$f(x,y)=\left(\frac 3{2y}-\frac yx\right)^2+\left(\frac 1y-\frac yx\right)^2$$ for all $(x,y)$ such that $xy\neq 0$. For such $(x,y)$, we have $f(x,y)=f(x,-y)$ so we may assume that $y<0$. If we don’t have $\displaystyle\lim_{\substack{(x,y)\to (0,0)\\ y<0}} f(x,y)=+\infty$, there exists an $A>0$ and two sequences $(x_n)$ and $(y_n)$, with $y_n<0$ for all $n$, such that $\lim x_n=\lim y_n=0$ and $f(x_n,y_n)<A^2$ for all $n$. Hence, for all $n$, $$\left(\frac 3{2y_n}-\frac{y_n}{x_n}\right)^2<A^2$$ and $$\left(\frac 1{y_n}-\frac {y_n}{x_n}\right)^2<A^2$$ From these two inequalities, we get for all $n$: $$-A<\frac 3{2y_n}-\frac{y_n}{x_n}<A$$ $$-A-\frac 1{2y_n}<\frac 1{y_n}-\frac {y_n}{x_n}<A$$ $$-\frac 1{2y_n}<\frac 1{y_n}-\frac {y_n}{x_n}+A<2A$$ But $$\lim -\frac 1{2y_n}=+\infty$$, hence this sequence can’t be bounded from above by $2A$. It proves that $$\lim_{\substack{(x,y)\to (0,0)\\ y<0}} f(x,y)=+\infty$$, hence the result.