Let f(x)=frac{1}{x-3}. To say that
lim_{x->oo}f(x)=0
means that f(x) can be made as close as desired to 0 by making the independent variable x close enough to oo.
Let the positive number varepsilon be how close one wishes to make f(x) to 0. Let delta be a real number that denotes how close one will make x to oo.
The limit exist if, for every varepsilon>0, there exist a delta\inRR such that
0-varepsilon<\f(x)<0+varepsilon
for all x>delta.
We already know that f(x)>0>0-varepsilon for all x>3. All that is left is the upper bound.
f(x)<\varepsilon
The inequality can be simplified to
x>\frac{1}{varepsilon}+3
Let delta=frac{1}{varepsilon}+3. We can see that for all x>delta(>3),
f(x)=frac{1}{x-3}<\frac{1}{delta-3}=varepsilon
