*The limit of a removable discontinuity is simply the value the function would take at that discontinuity if it were not a discontinuity.*

For clarification, consider the function #f(x) = (sin(x))/x#. It is clear that there will be some form of a discontinuity at #x=1# (as there the denominator is 0). **L'hospital's Rule** states, in part, that for a function #f(x) = g(x)/(h(x))#, where both are differentiable and #h(x)!=0# on an open interval containing #c# (except possibly at #c# itself), if the limits of both functions approach 0 or if both of the functions approach #+-oo# as #x->c#, the following can be stated.

#lim_(x->a) (g(x))/(h(x)) = lim_(x->a) (g'(x))/(h'(x))#

In this case, using L'hospital's Rule with #g(x) = sin(x), h(x) = x# yields #g'(x) = cos(x), h'(x) = 1.# Thus, the limit of our original #f(x)# as #x->0# will simply be #cos(0)/1#, or 1. Notice that if the function were not discontinuous at #x=0#, this is exactly the value that #f(x)# would have taken.