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.

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\begin{align}
{\ln\pars{\cosh\pars{x}} \over \ln\pars{\cos\pars{x}}} &
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{\ln\pars{1 + x^{2}/2} \over \ln\pars{1 - x^{2}/2}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{x^{2}/2 \over -x^{2}/2} = \bbox[15px,#ffc,border:1px groove navy]{1}
\end{align}
