TITLE: How to calculate $\lim_{x \to 0} \frac{e^{\tan^3x}-e^{x^3}}{2\ln (1+x^3\sin^2x)}$?
QUESTION [3 upvotes]: 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.

REPLY [2 votes]: 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.
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