El_Arrow
  • El_Arrow
need help with limit problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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El_Arrow
  • El_Arrow
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El_Arrow
  • El_Arrow
okay so my question is why is it dividing by n?
El_Arrow
  • El_Arrow
i thought you divided by the value with the largest power

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Melodious
  • Melodious
guys pls help me out in my question
El_Arrow
  • El_Arrow
@amistre64
El_Arrow
  • El_Arrow
@IrishBoy123
ganeshie8
  • ganeshie8
Next you should be asking why is it dividing by the term with largest power
El_Arrow
  • El_Arrow
i dont understand
ganeshie8
  • ganeshie8
Notice that "n" is indeed the largest power in the denominator, but clearly you're missing the key point, dividing by "n" is not really necessary here
ganeshie8
  • ganeshie8
Look at the expression \[\dfrac{n^3}{n+8}\] what happens to this term as "n" becomes large ? which one grows faster, numerator or denominator ?
El_Arrow
  • El_Arrow
the denominator
ganeshie8
  • ganeshie8
are you saying "n+8" grows faster than "n^3" ?
ganeshie8
  • ganeshie8
plugin n=10, 100 etc and see which one is growing faster
El_Arrow
  • El_Arrow
no i meant the numerator
ganeshie8
  • ganeshie8
Okay what about the value of expression as "n" gets large ?
El_Arrow
  • El_Arrow
it goes to infinity right
ganeshie8
  • ganeshie8
|dw:1436635975751:dw|
ganeshie8
  • ganeshie8
As you can see the function f(x) = x^3/(x+8) is increasing without any bound as x increases, so the corresponding sequence diverges
ganeshie8
  • ganeshie8
They are dividing top and bottom by "n" so that it becomes easy for you to see the same
ganeshie8
  • ganeshie8
\[a_n = \dfrac{n^3}{n+8} = \dfrac{n^2}{1+8/n}\] "plugin" \(n = \infty\), the expression becomes \[ \dfrac{\infty^2}{1+8/\infty} = \dfrac{\infty^2}{1+0} = \infty\]
El_Arrow
  • El_Arrow
i think i understand better now
ganeshie8
  • ganeshie8
In general : For any rational function, \(f(x) = \dfrac{P(x)}{Q(x)}\) , as \(x\to\infty\), we have \(f(x)\to\pm\infty\) if the degree of \(P(x)\) is greater than the degree of \(Q(x)\)

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