Anybody know if you can find limits approaching infinity without L'Hopital's rule?

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- precal

Anybody know if you can find limits approaching infinity without L'Hopital's rule?

- schrodinger

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- TuringTest

that depends on the limit

- precal

\[f(x)=4x(3x-\sqrt{9x^2+1})\]

- anonymous

yeah it depends if you can solve it without like the answer would be indeterminate forms like 0/0 then you would have to use L Hopitals rule

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## More answers

- precal

I am learning calculus but I have only been taught limits. I went to wolframalpha but it is using L Hopital's rule. I know that has to do with derivatives and I have not learn that yet.

- fj

De l'HÃ´pital rule is when the desire to limit the infinite

- anonymous

the answer here is -2/3= -0.66666666667

- anonymous

you can just use you calculator for that prob that you have then start with x=50 or 100 then 200 then 300..etc.

- precal

yes but my professor is BC (Before calculators). Therefore I have to do it by hand.

- anonymous

hmm i dont thnik that wolfram use L Hopitals rule in that particular problem. bec solving it by hand would take you more time to solved it some solution use here would be like solutions of Series Expansion..that would takes you to higher math.....

- anonymous

the best bet here to use here is your calculator then later on your calculus class your professor will teach you how to do this later on the lesson called Indeterminate forms maybe

- anonymous

another easy way for you to solve this is to graph them

- precal

graphing it how, just plotting points x=1, x=2, ...
Remember I am not allowed a calculator at all.

- Xishem

\[f(x)=4x(3x-\sqrt{9x^2+1})\]\[f(x)=12x^2-4x \sqrt{9x^2+1}\]\[f(x)=\frac{(12x^2-4x \sqrt{9x^2+1})(12x^2+4x \sqrt{9x^2+1})}{(12x^2+4x \sqrt{9x^2+1})}\]\[f(x)=\frac{144x^2-16x^2(9x^2+1)}{12x^2+4x \sqrt{9x^2+1}}\]I learned a rule in my high school calculus class that is based upon L Hopital's rule but doesn't actually require you to know how to use it.
1.) Determine what the highest order of x is in both the numerator and denominator
2.) There are three possible cases: The two orders are the same, the numerator's is higher, or the denominator's is higher.
If the numerator's order is higher, the limit is infinite.
If the denominator's order is higher, the limit is 0.
If the orders are the same, the limit is the ratio of the coefficients of the numerator and denominator.
In this case, after simplifying, you can determine what the highest coefficients in the numerator and denominator are:
\[f(x)=\frac{-16x^2}{24x^2}\]If you ignore the other variables, and expand the denominator, you'll find that the answer is:
\[\frac{-16}{24}=\frac{-2}{3}\]It's a pretty convoluted way to do it, especially when radicals are involved, but it works, and it's pretty easy to do without a calculator in this case.

- precal

Thanks I knew this was a way to do it but I forgot about using the conjugate. You are awesome :)

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