If #K = 0# then #g(x) = 0# for all #x!=c# in some open interval containing #c#, and so #lim_(x->c)(f(x)g(x)) = lim_(x->c)0=0#. Then, for the remainder of the explanation, we will assume #K > 0#.

Using the #epsilon-delta# definition of limits, we can say that #lim_(x->c)(f(x)g(x))=0# if for every #epsilon > 0# there exists a #delta > 0# such that #0<|x-c| < delta# implies #|f(x)g(x)-0| < epsilon#.

To show that this is the case, first we let #epsilon > 0# be arbitrary.

Now, as #K/epsilon>0# and #lim_(x->c)f(x)=0#, we know by the above definition there exists some #delta'>0# such that #|g(x)| < K# for #x in (c-delta',c+delta') \\\ {c}# and if #0<|x-c| < delta'# then #|f(x) - 0| < epsilon/K#.

Let #delta = delta'#. Then, for #0 < |x - c| < delta#, noting that as #|x-c| > 0# we know that #x!=c#, we have

#|f(x)g(x)-0| = |f(x)g(x)|#

#=|g(x)| * |f(x)|#

#<= K|f(x)|" "# (as #x in (c-delta',c+delta')\\\c => K >= |g(x)|#)

#< K(epsilon/K)" "# (as #0 < |x-c| < delta = delta' => |f(x)| < epsilon/K#)

#=epsilon#

Thus, as we have demonstrated the existence of such a #delta# for an arbitrary #epsilon#, we may conclude that #lim_(x->c)(f(x)g(x)) = 0#.