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Yet another limit

  1. Nov 13, 2004 #1
    Hi all, I can't find limit of this one:

    \lim \frac{(n + 4)^{100} - (n + 3)^{100}}{(n + 2)^{100} - n^{100}}

    I only got it to this point after I divided all expressions with n^100:

    \lim \frac{ \left( 1 + \frac{4}{n} \right) ^{100} - \left( 1 + \frac{3}{n} \right) ^{100}}{ \left( 1 + \frac{2}{n} \right) ^{100} - 1}

    I only can see that every expression goes to 1 in infinity, but I can't figure the limit out of this, anyway...

    Thank you for any suggestions. I would like to ask as well, what to do in cases like this - when I get [itex]\frac{0}{0}[/itex] or [itex]\frac{\infty}{\infty}[/itex] (and without l'Hospital).

    Thank you.
  2. jcsd
  3. Nov 13, 2004 #2

    matt grime

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    we can expand each bracket and write the quantity as:

    [tex]\frac{100n^{99} + P(n)}{200n^{99} + Q(n)}[/tex]

    where P and Q are polynomials of degree 98. Divide by n^{99} top and bottom and take the limit.
  4. Nov 13, 2004 #3
    Thank you matt, but could you please show me the way you got the expression you posted? I can't see how to expand the brackets...Thank you much.
  5. Nov 13, 2004 #4

    matt grime

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  6. Nov 13, 2004 #5
    Actually, that was the first I tried, but I didn't see the possibility to write the sums as a sum of two polynoms, one of which will go to zero when divided with n^99. Now I have it. Thank you for your time matt.
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