Limits which involve an explicit $0$

Question

I am trying to solve the following limit (or prove it doesn't exist)

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{||x,y||} $$

where $(x, y) \in R^2$.

I decided to analyze the limit over the y-axis, and reached the following expression

$$ \lim \frac{0}{|y|} $$

and I don't know what to do. Is this $0$, or an indeterminate form $\frac{0}{0}$? Should this be an indeterminate form, how do I proceed?

Is there an easier form to solve the previous limit?

Thanks in advance.

Answer 1

$$||x,y||=\sqrt{x^2+y^2}$$ And, $$0\le x^2\le x^2+y^2$$ $$0\le \frac{x^2}{\sqrt{x^2+y^2}} \le \sqrt{x^2+y^2}$$ $$ \lim_{(x,y) \rightarrow (0,0)}0\le\lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{||x,y||} \le \lim_{(x,y) \rightarrow (0,0)}\sqrt{x^2+y^2}$$ $$0\le \lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{||x,y||}\le 0$$ $$\Rightarrow \lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{||x,y||} = 0$$

For the other part of the question, $$\lim_{y \rightarrow 0} \frac{0}{|y|} = \lim_{y \rightarrow 0}0=0$$

This is because as per the definition of a limit, we can have $y$ arbitrarily close to $0$ but $y\ne 0$.

Answer 2

Use the generalized squeeze theorem:

$$0\le\frac{x^2}{\sqrt{x^2+y^2}}\le\frac{x^2}{\sqrt{x^2}}=\frac{x^2}{|x|}=\pm x\xrightarrow[x\to0]{}0$$