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#
