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Limit fraction

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  1. Jul 14, 2016 #1

    Rectifier

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    The problem
    [tex] \lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} [/tex]

    The attempt
    [tex] \lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{x \ln x}{x^4}) }{x + \left( \frac{2}{3} \right)^x} [/tex]


    Should I factor ## x ## or ## \left( \frac{2}{3} \right)^x ## in the denomenator?
     
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  3. Jul 14, 2016 #2

    malawi_glenn

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    what will happen to x+ (2/3)^x as x-> oo ?

    does x*ln(x) goes to oo faster or slower than x^4 as x-> oo ?
     
  4. Jul 14, 2016 #3

    Rectifier

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    I am aware that ##x^4## approaches infinity faster than the other expression in numerator. I am only interested in the denominator at this point. As for the first question ##x##
    approaches infinity while ##\left( \frac{2}{3} \right)^x## approaches 0.
     
  5. Jul 14, 2016 #4

    malawi_glenn

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    so your expression goes as x^4/x = x^3 as x->oo, right?
     
  6. Jul 14, 2016 #5

    Rectifier

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    Yup. But how come you factor x?
     
  7. Jul 14, 2016 #6

    malawi_glenn

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    what do you mean by "factor x"?

    I just note that x+(2/3)^x -> x as x->oo
     
  8. Jul 14, 2016 #7

    Rectifier

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    I meant that you factorize x from denominator. So we get ## \lim_{x\rightarrow \infty} \frac{x^4(1 + \frac{\ln x}{x^3}) }{x \left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) } = \lim_{x\rightarrow \infty} \frac{x^3(1 + \frac{ \ln x}{x^3}) }{\left( 1 + \frac{ \frac{2}{3} ^x}{x} \right) } ##.
     
  9. Jul 14, 2016 #8

    Ray Vickson

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    That would be one way to do it; another way would be to use l'Hospital's Rule; see, eg.,
    http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx
     
  10. Jul 16, 2016 #9

    micromass

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    What does that expression even mean??
     
  11. Jul 16, 2016 #10

    malawi_glenn

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    x + (2/3)^x has a linear asymptote with slope one as x -> oo

    is that better?
     
  12. Jul 16, 2016 #11

    micromass

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  13. Jul 16, 2016 #12

    malawi_glenn

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    Wasn't it obvious what I was trying to communicate?
     
  14. Jul 16, 2016 #13

    PeroK

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    Another technique is to estimate the numerator and/or denominator. In this case the numerator is greater than ##x^4## and the denominator is less than ##2x##, so the overall expression is greater than ##\frac{x^4}{2x}##.
     
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