Is it rigorous if the limit of a function depends on the variable? For example:
if $x$ is large enough, $\ln(1+x)$ approaches to $\ln(x)$ because
$$\lim_{x\to \infty} (\ln(1+x)-\ln(x))=0$$
Then Can we state that
$$\lim_{x\to \infty} \ln(1+x)=\lim_{x\to \infty} \ln(x)$$
Is it a rigorous statement or not? 
 A: Such statements are perfectly well formed, but it doesn't mean what you want it to. It merely says that those limits are equal - but both those limits are $\infty$. All the information that that tells us is both functions grow arbitrarily large - that is, we could equally well say
$$\lim_{x\rightarrow\infty}\ln(1+x)=\lim_{x\rightarrow\infty}e^x$$
even though those two functions grow at tremendously different rates. A limit is an operation that takes in a function and outputs a number. It discards all information about rate of convergence and whatnot (i.e. everything captured in the parameter $x$, which does not exist outside the limit), so it is not effective to express the notion of two functions getting arbitrarily close through two limits.
That said, we would be perfectly well justified in saying that $f$ and $g$ "approach each other" if
$$\lim_{x\rightarrow\infty}f(x)-g(x)=0.$$
However, it's not a perfectly natural notion - we could also say $f$ and $g$ do so if
$$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=1$$
which is a weaker notion, but still looks valid. We might only care
$$0<\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}<\infty$$
(which is the idea behind the very commonly used Big O notation). The idea is that the limit $\lim_{x\rightarrow\infty}f(x)-g(x)=0$ is expressing a very precise, but not wholly natural condition - so it's much better to write it out explicitly, as otherwise it is unclear exactly what you mean by "approach each other".
A: The first statement is true as it can be written in this form - 
$ \lim_{x \to \infty} log\left(\dfrac{1 + x}{x}\right)$
$ = \lim_{x \to \infty} log\left(1 + \dfrac{1}{x}\right)$
$ = log1 = 0$
