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Math Help - easy limit question.

  1. #1
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    easy limit question.

    Problem:


    Correct me if i am wrong, but i think i know that the delta x would be 2/n according to (b-a)/n which would also give 2i/n as our right interval. I would then factor out the 2/n and 2i/n and get 4/n^2. So i would end solving the limit of n -> infinity of
    \frac{4}{n^{2}}\left( \frac{n\left( n+1 \right)}{2} \right)^{2}

    trouble is im not sure how to evaluate this limit because the whole thing wrapped around the exponent is throwing me off.
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  2. #2
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    \sum {\left( {\frac{{2i}}<br />
{n}} \right)^3 \left( {\frac{2}<br />
{n}} \right) = \frac{{16}}<br />
{{n^4 }}\sum {i^3 } }
    \frac{{16}}<br />
{{n^4 }}\left( {\frac{{n(n + 1)}}<br />
{2}} \right)^2  = \frac{{16}}<br />
{{n^4 }}\left( {\frac{{n^4  + 2n^3  + n^2 }}<br />
{4}} \right) \to ?<br />
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  3. #3
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    Ahhh. I see. The limit would be 4. Cant believe i missed that. Anyways, i see I made a mistake. I do have one question for you. when you take the \left( \frac{2i}{n} \right)^{3} as a constant multiplied by the sum of f(x) why would the value of i be cubed if that too was a constant? im assuming its not but im not quite sure. thank you though.
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  4. #4
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    \left( {\frac{{2i}}<br />
{n}} \right)^3  = \left( {\frac{2}<br />
{n}} \right)^3 i^3
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