when given:
\begin{equation} \lim_{{(x,y) \to (0,0)}} f(x,y) = 0 \end{equation}
and if given equation that can be rewritten as:
\begin{equation} f(x,y) = \frac{V_1^2}{V_1^2 + V_2^2} \times something \end{equation}
Notice $\frac{V_1^2}{V_1^2 + V_2^2} < 1$
\begin{equation} \therefore f(x,y) = \frac{V_1^2}{V_1^2 + V_2^2} \times something < something \end{equation}
If I have $\sqrt{x^2+y^2} < \delta$
recall $|x|=\sqrt{x^2} \leq \sqrt{x^2+y^2} < \delta$ and $|y|=\sqrt{y^2} \leq \sqrt{x^2+y^2} < \delta$
which means $|ax| < a\delta$ and $|by| < b\delta$
try to match the $something$ use this inequality $|ax+by| < |ax| + |by| < a\delta + b\delta = (a+b)\delta$
then choose $\delta = \frac{\epsilon}{(a+b)}$
\begin{equation} f(x,y) = \frac{V_1^2}{V_1^2 + V_2^2} \times something < something < (a+b)\delta = (a+b)\frac{\epsilon}{(a+b)} \end{equation}
then I got $f(x,y) < \epsilon$ which is what I need.
Can I generalize it even more? For Example, can I do the same thing if the degree on the denominator is smaller than the degree on the numerator? Or if the limit is L.
BTW, is there any complete strategy guide for delta-epsilon proof I can study? Trying to compile a guide for the questions that can appear on an exam.