<s>[INST] 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?
 [/INST] 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
$$
\lim_{x\to0^+}\frac{\arctan(x^2)-\sin(x^3)-1+\cos x}{x^2}=
\lim_{x\to0^+}
\left(\frac{\arctan(x^2)}{x^2}-\frac{\sin(x^3)}{x^2}-\frac{1-\cos x}{x^2}\right)
$$
</s>