If we try to directly substitute #x = 0# into #lim_(x->0) x^(sinx)#, we end up with #0^0#, an indeterminate form. Thus, this is a case in which L'Hospital's Rule is useful.
Consider that our function is #y = x^(sinx)#. Taking the log of both sides gives #lny = ln(x^(sinx)) = sinx lnx#. L'Hospital's Rule generally works only on quotients, so we note that #sinx = 1 / (cscx)#.
This gives #lim_(x->0) lny = lim_(x->0) sinxlnx = lim_(x->0) lnx / (1/sinx)#. At this point we can actually apply L'Hospital's rule, giving: #lim_(x->0) lnx/(1/sinx) = lim_(x->0) (1/x)/(-cosx/(sin^2x)#.
We continue to rearrange and apply L'Hospital's rule as necessary. I will omit the #lim_(x->0)# for now.
#(1/x)/((-cosx)/(sin^2x)) = sin^2x/(-xcosx) => (L'H) => (2sinxcosx)/(-cosx + xsinx)#
At this point, we have a determinate form.
#lim_(x->0) (2sinxcosx)/(xsinx - cosx) = (2sin(0)cos(0))/((0)(sin(0)) - cos(0)#
#= 0 / (0 - 1) = 0/(-1) = 0#
So we've found that #lim_(x->0) lny = 0#. Note that #y = e^(lny)#. Thus our original limit can be written #lim_(x->0) e^(lny)#. Note that #e# is just a constant, so we can simply plug in the limit we found for #lny# as #x -> 0#, giving:
#lim_(x->0) x^(sinx) = lim_(x->0) e^(lny) = e^0 = 1#.
