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 $
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$\begingroup$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.
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