I would like to calculate limit: $ \lim_{n \to \infty}{n\cdot \ln({1-\arctan{\frac{3}{n}}})} $ I would like to calculate the following limit:
$$ \lim_{n \to \infty}{n\cdot \ln({1-\arctan{\frac{3}{n}}})} $$
We had this in an exam and I wasn't sure how to go about it. I guess I could try to play with it and use L'Hopitals, is there any easier way to go about it?
Any help would be greatly appreciated.
 A: Substitute the limit $n = {1\over x}$ and use L'Hospital's rule:
$$ \lim_{n \to \infty}{n\cdot \ln({1-\arctan{\frac{3}{n}}})} =\lim_{x \to 0}{{\ln({1-\arctan{3x}}})\over x}   =\lim_{x \to 0} {-3\cos^2(\arctan(3x))\over{{1-\arctan{3x}}}} = -3$$
A: Substitution Approach
Let $u=\arctan\left(\frac3n\right)$, then as $n\to\infty$, $u\to0$. Furthermore, $n=\frac3{\tan(u)}$. Then
$$
\begin{align}
\lim_{n\to\infty}n\log\left(1-\arctan\left(\frac3n\right)\right)
&=\lim_{u\to0}3\frac{\log(1-u)}{\tan(u)}\\[6pt]
&=3\frac{\lim\limits_{u\to0}\frac{\log(1-u)}{u}}{\lim\limits_{u\to0}\frac{\tan(u)}{u}}
\end{align}
$$
Then use common limits for $\frac{\log(1-u)}{u}$ and $\frac{\tan(u)}{u}$
A: Using the equivalences $\ln(1+t)\underset{t\to{0}}{\sim}t$ and $\arctan{t}\underset{t\to{0}}{\sim}t,$ we have $$\ln\left({1-\arctan{\frac{3}{n}}}\right)\underset{n\to{\infty}}{\sim} -\frac{3}{n},$$herefore, $$\lim\limits_{n \to \infty}{n \ln\left({1-\arctan{\frac{3}{n}}}\right)}=-3.$$
A: It goes as follows:
$$\lim_{n\to\infty} n\cdot \ln \left(1-\arctan \frac{3}{n}\right)$$
$$=\lim_{u\to 0} \frac{1}{u}\cdot \ln \left(1-\arctan 3u\right)$$
$$=\lim_{u\to 0} \frac{\ln \left[1+(-\arctan 3u)\right]}{(-\arctan 3u)} \cdot \frac{(-\arctan 3u)}{3u}\cdot 3$$
$$=1\cdot (-1)\cdot 3 = -3$$
