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Math Help - limit of a sequence

  1. #1
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    limit of a sequence

    Here's another limit:

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

    That's all I could get from it:

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

    And everything would be quite fine if there was no 2^{n^2}.... Then the limit would be e^\frac{3}{2}, just like the answers say.... Am I wrong somewhere, or is there a mistake in this example?

    Thank you in advance.
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  2. #2
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    That expresion is equivalent to:
    \left[ {\frac{{\left( {1 + \frac{2}{{n^2 }}} \right)^{n^2 } }}{{\left( {1 + \frac{1}{{2n^2 }}} \right)^{n^2 } }}} \right]\left[ {\frac{1}{{2^{n^2 } }}} \right]

    What is the limit of a product?
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  3. #3
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    It's equal to the product of the limits.... But the limit of \frac{1}{2^{n^2}} when n tends to infinity is zero, I believe; so the final result will be zero too?
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  4. #4
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    Quote Originally Posted by disclaimer View Post
    But the limit of \frac{1}{2^{n^2}} when n tends to infinity is zero, I believe; so the final result will be zero too?
    Don't just believe it. Know that it is true.
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  5. #5
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    Thank you. Now I guess that there's a mistake in my textbook, that's why I was a bit unsure. Thanks for clearing it up.
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  6. #6
    GAMMA Mathematics
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    The bottom is always more than the top and it grows faster, so the limit goes to zero.
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