Prove a limit with $\varepsilon$-$\delta$ definition:

Question

the limit we want to prove is: $$\lim_{(x,y)\to(0,0)}\frac{x^4}{x^2+y^2}=0$$ It is actually very simple: the denominator is the square of the norm of $(x,y)$, thus it follows that, for every $\varepsilon>0$ we must find $\delta>0$ for which it holds: $$\frac{x^4}{\delta^{2}}<\varepsilon$$ but I do not see how to get rid of that $x^4$ in order to define a proper $\delta$. I imagine one has to use some particular inequality based on squares I must have forgot.

Answer 1

$$0\le \frac{x^4}{x^2+y^2}=x^2\frac{x^2}{x^2+y^2}\le x^2\le x^2+y^2=\|(x,y)\|^2$$ So just take $\delta=\sqrt{\varepsilon}$

Answer 2

You can use the fact that$$\frac{x^4}{x^2+y^2}\leqslant\frac{x^4+2x^2y^2+y^4}{x^2+y^2}=x^2+y^2=\|(x,y)\|^2.$$So, for each $\varepsilon>0$; you can simply take $\delta=\sqrt\varepsilon$.

Answer 3

Alternative approach:

For any $(x,y)$, the norm $|x,0| \leq $ the norm $|x,y|$.

Therefore, $x^2 = |x,0|^2 \leq |x,y|^2 = \delta^2.$

Therefore, $x^4 \leq \delta^4.$

Thus, $\displaystyle \frac{x^4}{\delta^2} \leq \frac{\delta^4}{\delta^2} = \delta^2.$

Therefore, if you set $\delta < \sqrt{\epsilon}$, you will have $\delta^2 < \epsilon.$

This will imply that $\displaystyle \frac{x^4}{\delta^2} < \epsilon.$