TITLE: When to use L'Hospital's rule vs the limit shortcut
QUESTION [5 upvotes]: Before being so quick to downvote or throw darts, please forgive my ignorance and inability to recall basic calculus atm.
Consider the limit $\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}$. It can quickly be determined that the $x$ approaches $3$ using L'Hospital's rule.
The same answer can be derived, however, by using this shortcut (I am unaware if it has a specific name):
$\lim \limits_{x \to \infty}\frac{3x^2+14x-5}{x^2+x-12}=\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}$
Every term is divided by the highest degree of $x$ in the denominator. The terms with $x^2$ in both the numerator and denominator simplify. All other terms go to $0$. Therefore:
$\lim \limits_{x \to \infty}\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}=\frac{3}{1}=3$
1) What are the explicit conditions in which may I use this shortcut? 
2) How do I know when to use it over L'Hospital's rule, as both techniques can be used only when working with quotients?
3) What is the shortcuts name, if any?

REPLY [3 votes]: In general, suppose you have an expression of the form
$$f_1(x)+f_2(x)\over g_1(x)+g_2(x)$$
for which 
$$\lim_{x\to c}{f_2(x)\over f_1(x)}=\lim_{x\to c}{g_2(x)\over g_1(x)}=0$$ Then
$$\lim_{x\to c}{f_1(x)+f_2(x)\over g_1(x)+g_2(x)}=\lim_{x\to c}{f_1(x)\over g_1(x)}$$
The proof amounts to inserting the intermediate expression
$$\lim_{x\to c}{f_1(x)\over g_1(x)}\cdot{1+{f_2(x)\over f_1(x)}\over1+{g_2(x)\over g_1(x)}}$$
A nice example, in which L'Hopital gets you nowhere, is
$$\lim_{x\to\infty}{5e^{3x}+2e^x\over2e^{3x}+7e^{2x}}=\lim_{x\to\infty}{5e^{3x}\over2e^{3x}}=\lim_{x\to\infty}{5\over2}={5\over2}$$
The trick, in general, is to recognize a dominant term in the numerator and/or denominator. In essence, the shortcut says you can ignore all the other stuff. But you have to make sure that you pick out terms that really do dominate the other stuff; sometimes you can eyeball it, and sometimes you make mistakes (at least I do).