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.
