Do the limits exist? Determine whether the following limits exist and determine them in case of convergence:
1.) 
$$
\lim_{n\to\infty}\frac{\arctan(\exp(-n))}{\exp(n^2)}-\frac{\ln(n+1)}{n}.
$$
2.)
$$
\lim_{x\downarrow 0}\frac{\tan(x^2)-\ln(\ln(x+1))}{\ln(1/x)}
$$

Here's what I tried.
1.) Considering the first summand,
$$
\frac{\arctan(\exp(-n))}{\exp(n^2)}\to 0\text{ as }n\to\infty
$$
since the nominator tends to $\pi/2$ as $n\to\infty$ and the denominator tends to $\infty$ as $n\to\infty$. For the second summand, I apply L'Hospital, getting 
$$
\lim_{n\to\infty}\frac{\ln(n+1)}{n}=\lim_{n\to\infty}\frac{1}{n+1}=0.
$$
Both together, I get that that the searched limit exists and is 0.
2.) I am not that sure with this. I again consider two summands, namely 
$$
\frac{\tan(x^2)}{\ln(1/x)}\text{ and }\frac{\ln(\ln(x+1))}{\ln(1/x)}.
$$
Since $\tan(x^2)\to 0$ as $x\downarrow 0$ and $\ln(1/x)\to\infty$ as $x\downarrow 0$, the first summand should tend to $0$ as $x\downarrow 0$.
Considering the second summand,
$$
\ln(\ln(x+1))\to -\infty\text{ as }x\downarrow 0,
$$
and
$$
\ln(1/x)=+\infty\text{ as }x\downarrow 0.
$$
Hence, I am applying L'Hospital.
$$
\frac{d}{dx}(\ln(\ln(x+1)))=\frac{1}{(x+1)(\ln(x+1))}\to +\infty\text{ as }x\downarrow 0
$$
$$
\frac{d}{dx}(\ln(1/x))=-\frac{1}{x}\to -\infty\text{ as }x\downarrow 0
$$
So, I have to apply again L'Hospital, giving me
$$
\frac{d}{dx}\left(\frac{1}{(x+1)(\ln(x+1))}\right)=\frac{-1}{(x+1)^2\cdot\ln(x+1)}-\frac{1}{(x+1)^2\cdot (\ln(x+1))^2}\to -\infty, x\downarrow 0
$$
and
$$
\frac{d}{dx}\left(-\frac{1}{x}\right)=\frac{1}{x^2}\to +\infty, x\downarrow 0.
$$
Hm, another application of L'Hospital?

You are right about the first limit, with a slight note that $\arctan(\exp(-n))$ does not tend to $\pi/2$ as $n \to +\infty$ but rather to zero. I think you missed the minus in the exponent.
In the second limit, yes, you can use a second iteration of l'Hôpital's rule. That's a pretty common situation. One just needs to make sure they check the assumptions anew at each step, which you did. So go ahead taking a derivative of the new numerator and denominator and you'll find the numerical answer from there easily.
