Calculating limit of ln(arctan(x)) using chain rule

Question

I know there are other options to do that, but I'm trying to use the chain rule to calculate this limit and I get a wrong answer.

So what's wrong in the following calculation:

$\lim\limits_{x \to 0+} \ln(\arctan(x)) = \frac{1}{(1+x^2)} * \frac{1}{\arctan(x)} = \frac{1}{0+} = \infty $

Answer 1

Try this with the function $2x$.

$$\lim_{x\to 0} 2x =0.$$

Now take the derivative of $2x$, which is $2$:

$$\lim_{x\to 0} 2 = 2.$$

Hmmmm....different answers. This may be a L'hospital's Rule confusion.