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

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0that depends on the limit

precal
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=4x(3x\sqrt{9x^2+1})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah 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

precal
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0De l'Hôpital rule is when the desire to limit the infinite

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the answer here is 2/3= 0.66666666667

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can just use you calculator for that prob that you have then start with x=50 or 100 then 200 then 300..etc.

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes but my professor is BC (Before calculators). Therefore I have to do it by hand.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0another easy way for you to solve this is to graph them

precal
 4 years ago
Best ResponseYou've already chosen the best response.0graphing it how, just plotting points x=1, x=2, ... Remember I am not allowed a calculator at all.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=4x(3x\sqrt{9x^2+1})\]\[f(x)=12x^24x \sqrt{9x^2+1}\]\[f(x)=\frac{(12x^24x \sqrt{9x^2+1})(12x^2+4x \sqrt{9x^2+1})}{(12x^2+4x \sqrt{9x^2+1})}\]\[f(x)=\frac{144x^216x^2(9x^2+1)}{12x^2+4x \sqrt{9x^2+1}}\]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
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks I knew this was a way to do it but I forgot about using the conjugate. You are awesome :)
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