How to calculate the following limits: $\lim_{x\to0}\frac{\ln\left(\cosh\left(x\right)\right)}{\ln\left(\cos\left(x\right)\right)}$ 
How to calculate the following limits:
  
  
*
  
*$\lim\limits_{x\to0}\frac{\ln\left(\cosh\left(x\right)\right)}{\ln\left(\cos\left(x\right)\right)}$
  
*$\lim\limits_{n\to\infty}\sin\left(\pi\sqrt{n^{2}+1}\right)$

$1.$
$$\frac{\ln\left(\cosh\left(x\right)\right)}
{\ln\left(\cos\left(x\right)\right)}=\log\thinspace_{\cos\left(x\right)}\left(\cosh\left(x\right)\right)=\frac{
\log\thinspace_{\cos\left(x\right)}\left(\cosh\left(x\right)\right)}{\cosh\left(x\right)-
1}\cdot\left(\cosh\left(x\right)-1\right)$$
Also setting: $\cosh\left(x\right)-1=t$
 we have:
$$\lim_{t\to0}\frac{\log\thinspace_{\cos\left(x\right)}\left(t+1\right)}{t}t=\log\thinspace_{\cos\left(x\right)}\left(e\right).0=0$$
But this is not the answer, so where is my error?
$2.$
Based on my information about the properties of limits, since sine function is continuous over its 
interval hence the given limit can be rewritten as:
$$\sin\left(\pi\lim_{n\to\infty}\sqrt{n^{2}+1}\right)$$
which does not exist, but it's not the answer, so why I'm wrong?
Also if we consider the given function as a real valued 
function,e.g.$$\lim_{x\to\infty}\sin\left(\pi\sqrt{x^{2}+1}\right)$$ does not exist, so what is the reason 
behind this fact?
Why the limit of the function as a sequence does exist but as a real valued function we do not have such 
condition? 
any elementary hint for determining the first limit is appreciated.

Do you like an insane overkill? By the Weierstrass product for the cosine function
$$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right) $$
we have
$$ \cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right) $$
hence
$$ \frac{\log\cosh(x)}{\log\cos(x)}=\frac{\sum_{n\geq 0}\log\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right)}{\sum_{n\geq 0}\log\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right)}=\frac{-\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}}{-\sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^m 4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}}$$
and by switching the series on $n$ and $m$
$$ \lim_{x\to 0}\frac{\log\cosh(x)}{\log\cos(x)}=\frac{\sum_{n\geq 0}\frac{1}{(2n+1)^2}}{\sum_{n\geq 0}\frac{(-1)}{(2n+1)^2}}=\color{red}{-1}.$$
