Why can there be an infinite difference between two functions as x grows large, but a ratio of 1? I learned in grade school that the closer $a$ and $b$ are to one another, the closer $\frac{a}{b}$ is going to be to $1$. For example, $\frac{3}{\pi}$ is pretty close to 1, and $\frac{10^{100}}{42}$ isn't even close to 1.
So, why is:
$$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$$
But:
$$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$$ ?
Seems pretty counterintuitive. What's going on here?

$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$
As stated, this is because $x$ grows much more slowly than $x^2$. So the ratio goes to 1.
$1^2\over 1^2+1$$={1\over 1+1}$$={1\over 2}$$=0.5$
$2^2\over 2^2+2$$={2\over 4+2}$$={4\over 6}$$=0.66$
$3^2\over 3^2+3$$={9\over 9+3}$$={9\over 12}$$=0.75$
$10^2\over 10^2+10$$={100\over 100+10}$$={100\over 110}$$=0.91$
$100^2\over 100^2+100$$={10000\over 10000+100}$$={10000\over 10100}$$=0.99$  
$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$
This isn't a ratio, so the non-ratio can't go to 1. In fact, it's exactly equal to $x$ for any values:
$\lim_{x\to\infty}[(x^2+x)-(x^2)]$
$=\lim_{x\to\infty}[x^2+x-x^2]$
$=\lim_{x\to\infty}[x^2-x^2+x]$
$=\lim_{x\to\infty}[(x^2-x^2)+x]$
$=\lim_{x\to\infty}[(0)+x]$
$=\lim_{x\to\infty}[x]$
$=\infty$
Obviously, as $x$ goes to infinity, the limit diverges to infinity.
$1^2+1-1^2$$=1+1-1$$=1$
$2^2+2-2^2$$=4+2-4$$=2$
$3^2+3-3^2$$=9+3-9$$=3$
$10^2+10-10^2$$=100+10-100$$=10$
$100^2+100-100^2$$=10000+100-10000$$=100$  
