How to calculate $\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$? Question : 
$$\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$$
Here I tried $\tan x$ and $\sin x$ expansion in numerator and denominator which are as follows : 
$$\tan x =x +\frac{x^3}{3}+\frac{2x^5}{15} \cdots$$ 
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}\cdots$$ 
but this method is not working and also other alternative of using L'Hospital's rule is not working one this ... please help how to tackle this limit prblem thanks.

It looks like a bad guy. I used Wolfram Mathematica to compute
$$
\tan^3 x -x^3 = x^5+\frac{11}{15}x^7 + O(x^8)
$$
and 
$$
e^{\tan^3 x -x^3}=1+x^5+\frac{11}{15}x^7 + O(x^8).
$$
Therefore 
$$
e^{\tan^3 x}-e^{x^3}=x^5+\frac{11}{15}x^7 + O(x^8).
$$
Since
$$
\log(1+x^3 \sin^2 x) = x^5 +O(x^6),
$$
the limit exists and is equal to $1/2$.
But it can be very boring to perform these computations by hand: all the terms up to power $5$ cancel out, so you should take care of really many terms in each expansion, otherwise you'll get nothing useful.
