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precal

  • 4 years ago

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

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  1. TuringTest
    • 4 years ago
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    that depends on the limit

  2. precal
    • 4 years ago
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    \[f(x)=4x(3x-\sqrt{9x^2+1})\]

  3. anonymous
    • 4 years ago
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    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

  4. precal
    • 4 years ago
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    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.

  5. fj
    • 4 years ago
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    De l'Hôpital rule is when the desire to limit the infinite

  6. anonymous
    • 4 years ago
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    the answer here is -2/3= -0.66666666667

  7. anonymous
    • 4 years ago
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    you can just use you calculator for that prob that you have then start with x=50 or 100 then 200 then 300..etc.

  8. precal
    • 4 years ago
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    yes but my professor is BC (Before calculators). Therefore I have to do it by hand.

  9. anonymous
    • 4 years ago
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    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.....

  10. anonymous
    • 4 years ago
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    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

  11. anonymous
    • 4 years ago
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    another easy way for you to solve this is to graph them

  12. precal
    • 4 years ago
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    graphing it how, just plotting points x=1, x=2, ... Remember I am not allowed a calculator at all.

  13. Xishem
    • 4 years ago
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    \[f(x)=4x(3x-\sqrt{9x^2+1})\]\[f(x)=12x^2-4x \sqrt{9x^2+1}\]\[f(x)=\frac{(12x^2-4x \sqrt{9x^2+1})(12x^2+4x \sqrt{9x^2+1})}{(12x^2+4x \sqrt{9x^2+1})}\]\[f(x)=\frac{144x^2-16x^2(9x^2+1)}{12x^2+4x \sqrt{9x^2+1}}\]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.

  14. precal
    • 4 years ago
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    Thanks I knew this was a way to do it but I forgot about using the conjugate. You are awesome :)

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