Compute $\lim_\limits{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}$ I have to compute

$$
\lim_{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}
$$

I separated the numerator so I got that
$$\dfrac{-1+\cos(x)}{x\tan(x)+e^{x^3}-1} \longrightarrow -\dfrac{1}2;$$ 
I know that the limit is $\dfrac{1}2$ and I know, by checking on Wolfram Alpha, that
$$\dfrac{-\sin(x^3)}{x\tan(x)+e^{x^3}-1}\longrightarrow 0$$ so 
$$\dfrac{π/2- \arctan(1/x^2)}{x\tan(x)+e^{x^3}-1} \longrightarrow 1.$$
I tried using L'Hopital but it gets even more complicated. How can I solve it?
 A: Remember that, for $t>0$,
$$
\arctan t+\arctan\frac{1}{t}=\frac{\pi}{2}
$$
so we can write
$$
\frac{\pi}{2}-\arctan\frac{1}{x^2}=\arctan(x^2)
$$
This simplifies things a bit.
Now prove that
$$
\lim_{x\to0^+}\frac{x\tan x+e^{x^3}-1}{x^2}=1
$$
which follows from
$$
\lim_{x\to0^+}\frac{x\tan x}{x^2}=1
$$
and from
$$
\lim_{x\to0^+}\frac{e^{x^3}-1}{x^2}=
\lim_{x\to0^+}x\frac{e^{x^3}-1}{x^3}=0\cdot 1=0
$$
Thus the limit you have to compute is the easier
$$
\lim_{x\to0^+}\frac{\arctan(x^2)-\sin(x^3)-1+\cos x}{x^2}=
\lim_{x\to0^+}
\left(\frac{\arctan(x^2)}{x^2}-\frac{\sin(x^3)}{x^2}-\frac{1-\cos x}{x^2}\right)
$$
A: differentiate numerator
$d/dx(\pi/2 - \arctan(\frac{1}{x^2}) - \sin(\frac{1}{x^3})  - 1 + \cos(x))         $
$= -[(-2) / x^3] / [1 - (\frac{1}{x^2})^2] - 3x^2\cos(x^2) - \sin(x)$
$= 2x / [x^4 + 1] - 3x^2\cos(x^2) - \sin(x)$
you can differentiate that again, but i think that the only non zero term at x = 0 will be $2x / [x^4  +1] - \cos(x)$  - giving the value 1 for the numerator after two differentiations
for the denominator, the first differentiation is
$\tan(x) + x\sec^2(x) + 3x^2 \exp(x^3)$
a second differentiation of that, and setting x =0 is going to give the non-zero terms
$\sec^2(x) + \sec^2(x) + x(\sec(x) \tan(x)) + [zero terms]$
= $\sec^2(x) + \sec^2(x) = 2$
so I make it $\frac{1}{2}$
also I got it as 1/2 as an estimate on a computer, via numerical method
