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Math Help - Finding limit (basic question)

  1. #1
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    Finding limit (basic question)

    Lets say I have:
    <br />
\lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= ?<br />

    I know that the limit of of n, as n approaches infinity, is 2 but how I can get that right answer with the mathematical way? Of course I can quess answer with some big numbers but that's not what I'm looking for.

    Could someone please help me on this...
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  2. #2
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    \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}<br />

    \lim_{n\to\infty}\frac{2n^2 + 1 - 2n +2n}{n^2 - n}<br />

    \lim_{n\to\infty}\frac{2(n^2-n) +1+ 2n}{n^2 - n}<br />

    \lim_{n\to\infty} \left ( 2 +\frac{1+ 2n}{n^2 - n} \right )<br />

     2 + \lim_{n\to\infty}\frac{1+ 2n}{n^2 - n}<br />

     2 + 0 = 2 <br />
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  3. #3
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    Krizalid's Avatar
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    Quote Originally Posted by tabularasa View Post
    Lets say I have:
    <br />
\lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= ?<br />
    Why don't you divide top & bottom by n^2?
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  4. #4
    Senior Member polymerase's Avatar
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    Just divide to and bottom by the highest power, n^2

    \lim \frac{2+\frac{1}{n^2}}{1-\frac{1}{n}} then when the limit goes to infinity the fractions go to zero....

    Hence, \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= \lim_{n\to\infty} \frac{2+\frac{1}{n^2}}{1-\frac{1}{n}}=2
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