Question

Applying L'Hopital's Rule: (Problem below).

Answer 1

Find \[\lim_{x \rightarrow \infty}\frac{ 2x ^{2}+3x }{ x ^{3}+x+1 }\]
I always check my problems using derivative calculators or Mathematica to see if i've done them right, and since Mathematica usually gives exact instructions on how they did it, that usually solves the problem itself. However, since it's asking to explicitly apply L'Hopital's rule (I know that all of these can be solved using other methods), nothing really does that, or at least, with steps in between. What i'm ending up with is:
\[\lim_{x \rightarrow \infty}\frac{ 2x ^{2}+3x }{ x ^{3}+x+1 } = \lim_{x \rightarrow \infty}\frac{ 4x+3 }{ 3x ^{2}+1 }\]At this point, i'm not sure if I have to continue, because I feel like the greater degree of the denominator would just lead to zero and that would be okay, because it's not *actually* zero, it's just infitesimally small to the point at which the value is for all practical purposes zero. I could continue on like so:\[ \lim_{x \rightarrow \infty}\frac{ 4x+3 }{ 3x ^{2}+1 } = \lim_{x \rightarrow \infty} \frac{ 4 }{ 6x }\]And that seems more reassuring by the fact that there simply isn't a value approaching infinity in the numerator, but i'm not sure if that's taking L'Hopitals rule too far and using it past the point where the limit is determinate.

Answer 2

You could do this w/o l'Hopital, if you divide (in the first step) everything by x³, you will get:\[\lim_{x \rightarrow \infty}\frac{ \frac{ 2 }{ x }+\frac{ 3 }{ x^2 } }{ 1+\frac{ 1 }{ x^2 }+\frac{ 1 }{ x^3 } }=\frac{ 0+0 }{ 1+0+0 }=0\]

Answer 3

You have taken l'Hopital as far as it gets. Once you've arrived at 4/(6x), l'Hopital no longer applies, because the numerator is constant (4) and the denominator is going to infinity.

So you might as well do the divide by x³ trick right at the beginning.

Answer 4

I know that I could have, I said that in the problem, but i'm supposed to do it with L'Hopital's.

Answer 5

So would (just pretending) \[\frac{ 4x }{ 3x ^{2} }\] not be acceptable because even though the degree of the denominator is greater, it's still infinity and infinity would scrap that presumption?

Answer 6

It would be acceptable for l"Hopital, because 4x and 3x² go to infinity.
It would give you 4/(6x), or 4/6 * 1/x. That 1/x is a standard limit (0), so you're done.

My only point is: by just dividing by the highest power of x, you'd get that same standard limit, so using l'H is a bit over the top imo. I'd save l'Hopital for situations I wouldn't know how to handle without it.

Then again: you were "forced"to use it...

Answer 7

No, not for applying L'Hopital's rule, i'm saying leaving the limit as is. Will get back to this later, going out. Yes, I was, because that was the problem prompt. I'm not going to ignore directions. Sorry that apparently that's an issue.

Answer 8

Not an issue, just my opinion :)
Thx.

Answer 9

Okay, made sense of my fundamental question through other meanings. Regardless of the degree of a polynomial with a variable, if you have a fraction with a term in the numerator and the term in the denominator with said variable, both will still be approaching infinity, even if the degrees are different. So you'll get infinity/infinity, and thus L'Hopital's rule can be applied. Thanks.