It's a school exercise with steps provided.
Find the limit of: $$ \lim_{(x_1,x_2)\rightarrow (0,0)} \frac{x_1x_2^2}{x_1^2-x_1x_2+x_2^2} $$
Solution:
$$ 0<||x||_2<\delta $$
$$ \bigg|\frac{x_1x_2^2}{x_1^2-x_1x_2+x_2^2}\bigg|\leqslant \bigg|\frac{x_1x_2^2}{0.5(x_1-x_2)^2+0.5(x_1^2+x_2^2)}\bigg|\leqslant 2||x||_2<2\delta $$
Then we have: $$ \delta =\frac{\varepsilon}{2} $$
$$ \lim_{x,y\rightarrow 0}f(x)=0 $$
Why the following part is true ? $$\bigg|\frac{x_1x_2^2}{0.5(x_1-x_2)^2+0.5(x_1^2+x_2^2)}\bigg|\leqslant 2||x||_2$$
I try to break each term and compare it with the norm.I am not sure if $|x_1x_2^2|\leqslant||x||^3$ is true and don't know how to calculate the other part.
You have$$\frac12(x_1-x_2)^2+\frac12(x_1+x_2)^2=x_1^{\,2}+x_2^{\,2}=\|x\|_2^{\,2}.$$On the other hand,$$|x_1x_2|=\sqrt{|x_1|^2|x_2|^2}\leqslant\frac{x_1^{\,2}+x_2^{\,2}}2=\frac{\|x\|_2^{\,2}}2.$$So,$$\left|\frac{x_1x_2^{\,2}}{\frac12(x_1-x_2)^2+\frac12(x_1+x_2)^2}\right|\leqslant\frac{|x_2|}2\leqslant\frac{\|x\|_2}2,$$which is a stronger inequality.