We may split the limit up as follows, recalling the fact that
#lim_(x->a)[f(x)+-g(x)]=lim_(x->a)f(x)+-lim_(x->a)g(x)#
Then,
#lim_(x->0^+)(1+1/x)=lim_(x->0^+)1+lim_(x->0^+)1/x#
#lim_(x->0^+)1=1,# in general, the limit to any value of a constant is simply that constant.
To determine #lim_(x->0^+)1/x,# envision dividing #1/x# by smaller and smaller positive numbers, as we're approaching #0# from the positive side:
#1/0.1=10#
#1/0.01=100#
#1/0.001=1000#
#1/0.0001=10000#
We can see as we approach #0# from the positive side, our result gets larger and larger, that is, heads toward #oo.#
Thus, #lim_(x->0^+)1/x=oo#
Our result is
#lim_(x->0^+)(1+1/x)=1+oo=oo,# the #1# doesn't change the fact that we're heading toward infinity, it doesn't impact the answer, #1# is insignificant in comparison.
