Evaluating $\lim_{x\to0} \frac{ax-\arctan(ax)}{bx-\arctan(bx)}$ I struggle with solving this with L'Hospital technique:

$$\lim_{x\to0} \frac{ax-\arctan(ax)}{bx-\arctan(bx)} \tag{1}$$

It should be:
$$\lim_{x\to0} \dfrac{a-\dfrac{a}{a^2x^2+1}}{b-\dfrac{b}{b^2x^2+1}} \tag{2}$$
I end up with:
$$\frac{a-a/1}{b-b/1} \tag{3}$$
Which is $0/0$.
Just the same as if I didn't use l'Hospital method.
Same if "tan" is used instead of "arctan".
Yet, it seems it is solvable.
Any clue in order to get me on the right path?
Thanks.
 A: Hint:
Use 
$$a-\dfrac a{a^2x^2+1}=\dfrac{a^3x^2}{a^2x^2+1}$$
Alternatively 
Using Are all limits solvable without L'Hôpital Rule or Series Expansion
$$\lim_{x\to0}\dfrac{ax-\arctan(ax)}{(ax)^3}=\dfrac13$$
A: $\lim_{x \to 0} \frac {ax - \arctan (ax)} {bx - \arctan (bx)}=\lim_{x \to 0} \frac {a-a/(1+a^{2}x^{2})} {b-b/(1+b^{2}x^{2})}$. Simplify this ratio.  The limit is $\frac {a^{3}} {b^{3}}$ since the ratio  simplifies to $\frac {a^{3}} {b^{3}} \frac {1+b^{2}x^{2}} {1+a^{2}x^{2}}$ after cancelling $x^{2}$.
A: We have
$$
\lim_{x\to0} \dfrac{ax-\arctan(ax)}{bx-\arctan(bx)} = \lim_{x\to0}\frac{\sin\left(ax-\arctan(ax)\right)}{\sin\left(bx-\arctan(bx)\right)} = \lim_{x\to 0}\frac{\sin(a x)-a x \cos(ax)}{\sin(b x)-a x \cos(b x)}\frac{\sqrt{1+b^2x^2}}{\sqrt{1+a^2x^2}}
$$
now discarding $\frac{\sqrt{1+b^2x^2}}{\sqrt{1+a^2x^2}}$ we follow with
$$
\lim_{x\to 0}\frac{\sin(a x)-a x \cos(ax)}{\sin(b x)-a x \cos(b x)}
$$
Applying l'Hopital twice now, we obtain easily as result
$$
\frac{a^3}{b^3}
$$
A: You just keep on applying L'Hopital's rule. After three times you will get$$\lim_{x\to0}\frac{ax-\arctan(ax)}{bx-\arctan(bx)}=\frac{2a^3}{2b^3}=\left(\frac ab\right)^3.$$
