Rearranging Limit Problem Having some trouble expressing the following:
Find $\displaystyle \lim_{x\to\infty} \frac{2-3x+4x^2}{7 + 12x + 3x^2}$, by first expressing the limit as $\displaystyle \lim_{x\to0} f(x)$, for some real function $\displaystyle f$
My initial thoughts would be to switch the numerator and the denominator and then evaluate at $\displaystyle x\to0$, which would leave an answer of $\displaystyle \frac{7}{2}$, but this seems too easy and is thus more than likely incorrect.
 A: First write
$$
{2-3x+4x^2\over 7+12x+3x^2}= 
{{2\over x^2}-{3\over x}+4\over {7\over x^2}+{12\over x}+3}.
$$
Then the limit is expressed by (setting $u=1/x$ in the above)
$$
\lim_{u\rightarrow0}{{{2u^2}-{3u}+4\over {7u^2}+{12u}+3}} .
$$
A: Changing variables is inefficient. Rather, simply employ the useful Principle of Domination to deduce the that limit of a quotient is the limit of the quotient of the "dominant" terms, namely factor out the dominant terms from each sum as follows (where limits are as $\rm\:x\to \infty)$
$$\rm\ \frac{f}g\to a,\ \ \frac{f_j}f \to 0,\ \ \frac{g_j}g \to 0\ \ \Rightarrow\ \ \frac{f + f_1 +\cdots f_n}{g + g_1\cdots g_k} \ =\ \frac{f}g\: {\frac{1 + \dfrac{f_1}f+\cdots+\dfrac{f_n}f}{1 + \dfrac{g_1}g +\cdots + \dfrac{g_k}g}}\:\!\to\: a $$
Therefore $\rm\ \ \dfrac{x}{x^2} = \dfrac{1}{x}\to 0,\ \ \dfrac{1}{x^2}\to 0\ \ \Rightarrow\ \ \dfrac{ax^2+bx+c}{dx^2+ex+f}\ =\ \dfrac{a + \dfrac{b}x + \dfrac{c}{x^2}}{d + \dfrac{e}x + \dfrac{f}{x^2}}\to\: \dfrac{a}d $
Should you go on to study the algebraic approach to limits and asymptotics via valuation theory you will find that an analogous principle of domination plays a key role.
