$\begingroup$Limits that involve infinity are not a problem for me, but there is an idea that is used in rational limits that I never really understood why.
The so called "rule" says that given a rational expression, if you want to find the limit as $x$ goes to infinity, just find the highest degree in the denominator and divide every term by it.
Consider the following example :
$\displaystyle\lim_{x \to \infty}\frac{3x^3 + 5x - 2}{2x^2 + 1}$.
I will not write all the work out, but in this example, $x^2$ is the highest degree in the denominator so I divide every term by $x^2$. This will essentially make each term go to zero except $3x$, and so when you put infinity in you get infinity.
My question is, why are we specifically finding the highest degree in the denominator? Is there a particular reason? Why couldn't we do the highest degree in the numerator or even the highest degree in the whole expression? Actually whats interesting is if you divide everything by the highest degree in the numerator your left with just $3$ as an answer which is incorrect so there must be a logical reason for this so called rule.
$\endgroup$13 Answers
$\begingroup$Considering expressions such as $$y=\frac{3x^3 + 5x - 2}{2x^2 + 1}$$ you could use long division and get $$y=\frac{3 x}{2}+\frac{7}{4 x}+\cdots$$ which shows the slant asymptote and how the function approaches it.
Otherwise, you could factor the highest degree in both numerator and denominator to write $$y=\frac{x^3\left(3+\frac 5 x-\frac 2{x^3}\right)}{x^2\left(2+\frac 1{x^2}\right)}=x\frac{\left(3+\frac 5 x-\frac 2{x^3}\right)}{\left(2+\frac 1{x^2}\right)}$$ Since $x\to \infty$, all terms with $x$ in denominator becomes negligible and you just keep the constants.
$\endgroup$$\begingroup$If you want to do this "rigorously," you don't just divide the $2x^2$. You divide the whole polynomial into the numerator to get: $$\lim_{x \to \infty} \frac{3}{2}x+\frac{\frac{7x-4}{2}}{2x^2+1}$$ Now, we look at the biggest term, which is $\frac{3}{2}x$. This goes to $\infty$ as $x\to \infty$. However, in the other part, the denominator grows faster than the numerator and thus it goes to $0$ as $x \to \infty$. Thus, the answer is $\infty+0=\infty$.
This is where this idea of dividing just by $2x^2$ comes from. We get the same result as if we did it the correct way, by dividing the numerator by $2x^2+1$, but we get the correct answer much faster. All that matters is the biggest term since everything else is ignored since it's small compared to the biggest term, so we only have to look at the biggest term in order to get the answer to the limit. Either way, we get the same biggest term of $\frac{3}{2}x$ term that goes to $\infty$, so our answer is $\infty$, so it works in both methods, so we might as well use the easier method of dividing by $2x^2$.
$\endgroup$3$\begingroup$You can just as well divide by the largest power of $x$ in the numerator. In your example $f(x)=(3 x^3+5 x-2)/(2 x^2+1) ,$ when $x\ne 0$ we have $$f(x)=(3+5 x^{-2}+x^{-3})/(2 x^{-1}+x^{-3}).$$ As $x\to \infty$ that numerator converges to $3$ and that denominator converges to $0$ through positive values, so $f(x)\to +\infty.$
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