The keys to solving this are to take the logarithm of the given function then use Taylor series centered at 0 (Maclaurin series) to determine how the expressions #[f(x)]# and #{f(x)}# can be simplified when #x# is close to zero. Then use Taylor series again to calculate the resulting limit before exponentiating to get the final answer.
Let #f(x)=tan(x)/x#, let #g(x)=([f(x)]+x^2)^{1/{{f(x)}}}#, and let
#h(x)=ln(g(x))=(ln([f(x)]+x^2))/({f(x)})#, where the last equality follows by a property of logarithms.
Now when #x# is close to zero (within 1 unit of 0 is good enough in this case), then #[f(x)]=[tan(x)/x]=1# since #1 < tan(x)/x < 2# when #-1 < x < 1# (also see the Taylor series expansions below).
When #-1 < x < 1# and #x!=0# we also have #{f(x)}={tan(x)/x}=tan(x)/x-1=x^2/3+2/15 x^4+cdots#. The reason for this follows from the Taylor expansion for #f(x)=tan(x)#, which starts #tan(x)=x+x^3/3+2/15 x^5+cdots#.
This implies that #tan(x)/x=1+x^2/3+2/15 x^4+cdots# for #-1 < x < 1# when #x!=0# (actually it works for #-pi/2 < x < pi/2# when #x!=0#).
Certainly then, when #x# is sufficiently close to zero it follows that #{tan(x)/x}=x^2/3+2/15 x^4+cdots#. But this is equal to #tan(x)/x-1# for #x# sufficiently close to zero.
The Taylor series for #ln(1+x^2)# is #ln(1+x^2)=x^2-x^4/2+x^6/3-x^8/4+cdots# for #-1 < x < 1#.
Hence, when #x# is sufficiently close to zero, #h(x)=ln(1+x^2)/(tan(x)/x-1)=(x^2-x^4/2+cdots)/(x^2/3+2/15 x^4+cdots)#.
As #x->0#, the dominant terms are those of the lowest powers, so #lim_{x->0}h(x)=1/(1/3)=3# (L'Hopital's Rule could also be used twice here).
Therefore, #lim_{x->0}g(x)=lim_{x->0}e^{h(x)}=e^{lim_{x->0}h(x)}=e^{3}#.
