<s>[INST] Let $R(x) := \frac {3x^4+5x+2} {2x^4+3x^2+7}$. Prove $R$ has a limit for $x \rightarrow \infty$. Let $$R(x) := \frac {3x^4+5x+2} {2x^4+3x^2+7}$$
Prove $R$ has a limit for $x \rightarrow \infty$.
Using a calculator I've expaneded $\frac {3x^4+5x+2} {2x^4+3x^2+7} = \frac {-9x^2}{2(x^4+3x^2+7)} +\frac {5x}{2x^4+3x^2+7} -  \frac {17}{2(x^4+3x^2+7)} + 3/2$.
And I can easily see that three of these fractions has limit $0$ for $x \rightarrow \infty$.
However am I relying a bit to much on a calculator here ? And should I prove for each of these fractions rigorous that the limit is $0$ ? 
 [/INST] There's a simpler way:
$$
\lim_{x\to\infty}\frac{3x^4+5x+2}{2x^4+3x^2+7}=\lim_{x\to\infty}\frac{3+\frac5{x^3}+\frac2{x^4}}{2+\frac3{x^2}+\frac7{x^4}}=\frac32
$$
since
$$
\lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty} g(x)}
$$
and
$$
\lim_{x\to\infty}(f(x)+g(x))=\lim_{x\to\infty}f(x)+\lim_{x\to\infty}g(x)
$$ provided that both limits exist and $\lim_{x\to\infty}g(x)\ne0$ in the former case.
</s>