limit and convergence I have two functions, $f(x)=x$ and $g(x)=x-1$ 
$$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$$
this means that when $x$ goes to infinity, those two function get closer and closer, So I think I can write it like this: 
$f(x)\approx g(x)$ as $x \rightarrow \infty$ but how come
$$\int^{f(x)}_{g(x)} 1 \, dy \approx 0$$
we know for shout that this function must be 1. I just don't understand limit says that two functions are almost same, but the integral gives different answer.
 A: The fact that 
$$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$$
does NOT mean that 
"when $x$ goes to infinity, those two function get closer and closer".  It only means that the two functions get PROPORTIONALLY close, in the sense that $\vert f(x) - g(x)\vert$ gets PROPORTIONALLY small when compared to $g(x)$. 
However, the absolute difference may $\vert f(x) - g(x)\vert$ may remain constant (or even increase). 
In the case of $f(x)=x$ and $g(x)=x-1$ the absolute difference $\vert f(x) - g(x)  \vert$ remains constant. So, in absolute terms, $f$ does not get closer to $g$.  
In the case of  $f(x)=x+\ln x $ and $g(x)=x$ the absolute difference $\vert f(x) - g(x)\vert$ increases but more slowly than $g(x)$. So, in absolute terms, $f$ even gets away from from $g$
Remark: Please note that it is easy to prove that
$$\int^{f(x)}_{g(x)} 1 \, dy = f(x)-g(x)$$
so your questions regarding the integrals are actually questions about $f(x)-g(x)$. 
A: Whilst I wouldn't really use the notation $f(x) \approx g(x)$, since a functions tail having similar values at the tail doesn't mean they are anywhere close to the same function. Humouring your result, however, it seems that both results you've put forward are consistent with one another.
You get, from the limit of the ratio that $$f(x) \approx g(x)$$ as $x \to \infty$ and your integral agrees with this, since $$\int^{f(x)}_{g(x)} 1 \, \mathrm{d}y \approx 0 \implies \bigg[y\bigg]_{g(x)}^{f(x)}\implies f(x) - g(x) \approx 0 \implies f(x) \approx g(x).$$
A: The integral says the same. Note that you can see that integral as a rectangle with sides 1 and $f(x)-g(x)$. The area of the rectangle is approximately 0, this means that  $f(x)-g(x) \approx 0$, and thus $f(x)\approx g(x)$
