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.