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A limit as x->inf

  1. Jan 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the limit of the expression below as x->inf

    2. Relevant equations

    [tex]\frac{\sqrt{x^2-4}(x^2 + 2)}{24x^3}[/tex]

    3. The attempt at a solution

    Well I know the answer is 1/24...

    I say use l'Hospitals rule but it gets messy, no? Is there a better way to do this?
     
  2. jcsd
  3. Jan 8, 2007 #2
    hint:Try multiplying and dividing the expression by "something" ;)
     
  4. Jan 8, 2007 #3
    Well I multiplied out the numerator then split it into two limits where l'Hospital's rule was easy on one of them, and the other was straightforward.

    What were you suggesting a use neutrino?
     
  5. Jan 8, 2007 #4

    AlephZero

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    Just write it as

    [tex]\frac{\sqrt{1 - 4/x^2 }(1 + 2/x^2)}{24}[/tex]
     
  6. Jan 8, 2007 #5

    HallsofIvy

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    [tex]\frac{\sqrt{x^2-4}(x^2 + 2)}{24x^3}[/tex]
    Has degree 3 in both numerator and denominator. What do you get if you divide both numerator and denominator by x3?
     
  7. Jan 8, 2007 #6

    Gib Z

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    I reckon AlephZero had the easiest way. Makes the answer obvious.
     
  8. Jan 8, 2007 #7
    Just replace x with inf. and discard any addition and subtraction in the polynomial. From there, it's just calculating endpoints. The equation becomes:

    (i^2^0.5 * i^2)/(24i^3) where i = infinity. Simplifying,
    (i*i^2)/(24i^3), which again simplifies to
    i^3/24i^3

    At this point, i^3 cancels eachother, leaving 1/24.
     
  9. Jan 8, 2007 #8

    Gib Z

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    why replace x with i? Your just using a different letter...
     
  10. Jan 9, 2007 #9
    Replace it with infinity, I just used 'i' as a quick forum substitute.
     
  11. Jan 9, 2007 #10

    dextercioby

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    JUst a piece of advice: always try to avoid using l'Hopital's rule when dealing with fractions containg (square, cubic,...) roots. There are cases when this rule doesn't lead you anywhere. :surprised Let me think of one and i'll post it later.

    Daniel.
     
  12. Jan 9, 2007 #11

    dextercioby

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    Yes, got it: Apply l'Hopital's rule to

    [tex] \lim_{x\rightarrow \infty} \frac{\sqrt{x^{2}+1}}{x} [/tex]

    and you'll understand my point better.

    Daniel.
     
  13. Jan 9, 2007 #12

    Gib Z

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    Yes, but in essence, you just substituted in infinity for x, except you did not carry out any operations on it until the end. eg You did not replace infinity square with infinity, you left it as it is. Your method works, but in essence its the same thing, and it could get confusing.
     
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