Limit fraction

  • #1
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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?
 

Answers and Replies

  • #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 ?
 
  • #3
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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 ?
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.
 
  • #4
malawi_glenn
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so your expression goes as x^4/x = x^3 as x->oo, right?
 
  • #5
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so your expression goes as x^4/x = x^3 as x->oo, right?
Yup. But how come you factor x?
 
  • #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
 
  • #7
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what do you mean by "factor x"?

I just note that x+(2/3)^x -> x as x->oo
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) } ##.
 
  • #8
Ray Vickson
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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) } ##.
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
 
  • #9
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I just note that x+(2/3)^x -> x as x->oo
What does that expression even mean??
 
  • #10
malawi_glenn
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What does that expression even mean??
x + (2/3)^x has a linear asymptote with slope one as x -> oo

is that better?
 
  • #12
malawi_glenn
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Wasn't it obvious what I was trying to communicate?
 
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  • #13
PeroK
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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?
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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