$\epsilon$-$\delta$ definition on $fg$

Question

Given that $$\lim_{(x,y)\to(a,b)}{f(x,y)} = l_1,\lim_{(x,y)\to(a,b)}{g(x,y)} = l_2$$ we need to prove that $$\lim_{(x,y)\to(a,b)}{f(x,y).g(x,y)} = l_1l_2$$ by the epsilon-delta definition.

Answer 1

One trick that is useful in cases like this is what I learned to call the "fat 0". The idea is to add and subtract the same value so as to turn it into something useful. In this case, consider: \begin{align} |f(x,y)g(x,y) - l_1l_2| &= |f(x, y)g(x, y) - f(x,y)l_2 + f(x,y)l_2 - l_1l_2| \\ &\leq |f(x, y)g(x,y) - f(x, y)l_2| + |f(x,y)l_2 - l_1l_2| \end{align} Now, the trick is that each of these summands should be something you can work with a bit better by using the existence of the limits for $f$ and $g$, respectively.

Hopefully this helps!