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Math Help - How do you find the limit of these sequences?

  1. #1
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    How do you find the limit of these sequences?

    a) (2n^2+n+sqrt(n)-3)/(n^2 -5n+7) as n tends to infinity
    b) (1+2+3+...+n)/n^2 as n tends to infinity
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  2. #2
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    For the first, we may divide the numerator and denominator by n^2. For the second, we may use the formula

    1+2+\cdots+n=\frac{n(n+1)}{2}.
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  3. #3
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    Quote Originally Posted by amm345 View Post
    a) (2n^2+n+sqrt(n)-3)/(n^2 -5n+7) as n tends to infinity
    b) (1+2+3+...+n)/n^2 as n tends to infinity

    \frac{2n^2+n+\sqrt{n}-3}{n^2-5n+7}=\frac{2n^2+n+\sqrt{n}-3}{n^2-5n+7}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}

    =\frac{2+\frac{1}{n}+\frac{1}{n^{3/2}}-\frac{3}{n^2}}{1-\frac{5}{n}+\frac{7}{n^2}}

    and you should be able to see now what the limit is....


    Always multiply by 1 in the special form of 1 over the highest term of the fraction, in fact, this idea leads to a rule about limits to infinity

    The highest power term always dominates, so if its on top then we get infinity, if it's on bottom we get 0, if the degree of top and bottom are equal, then the limit is the ratio of coefficients between the highest degree terms as is the case here

    If you want to see this argument rigorously, let me know
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  4. #4
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    I think I understand it and that was the limit I got the first time I tried, I just wanted to be sure, but if you don't mind seeing the rigorous version would help!
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  5. #5
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    Quote Originally Posted by amm345 View Post
    I think I understand it and that was the limit I got the first time I tried, I just wanted to be sure, but if you don't mind seeing the rigorous version would help!

    Alright, let's only deal with polynomials here...

    Say we have \lim_{n\rightarrow\infty}\frac{a_0n^{p_0}+a_1n^{p_  1}+...+a_n}{b_0n^{q_0}+b_1n^{q_1}+...+b_n}

    Now we might as well assume that  p_0>p_i for all other i since if it isn't, we just rearrange the expression on top to make it that way.... same on the bottom


    So now, 1 of 3 things can happen, either p_0=q_0, p_0>q_0 or p_0<q_0

    So case 1, what if p_0=q_0?

    \frac{a_0n^{p_0}+a_1n^{p_1}+...+a_n}{b_0n^{q_0}+b_  1n^{q_1}+...+b_n}=\frac{a_0n^{p_0}+a_1n^{p_1}+...+  a_n}{b_0n^{q_0}+b_1n^{q_1}+...+b_n}\cdot \frac{\frac{1}{n^{p_0}}}{\frac{1}{n^{p_0}}}

    =\frac{a_0+a_1n^{p_1-p_0}+...+\frac{a_n}{n^{p_0}}}{b_0+b_1n^{q_1-p_0}+...+\frac{b_n}{n^{p_0}}}

    Notice since we assume p_0=q_0 the cancellation happens on both the top and bottom. Also, since we assume both p_0 and q_0 are the largest exponents, everything left has a negative exponent, which means it is of the form c_i\frac{1}{n^y} where c_i and y are just 2 numbers. So the limit as n goes to infinity is zero in all those terms. So the limit is the only thing left which is \frac{a_0}{b_0}

    Now let's assume p_0>q_0

    Then we do the same thing but we multiply top and bottom by \frac{\frac{1}{n^{q_0}}}{\frac{1}{n^{q_0}}}

    This will cancel everything on the bottom or give it a negative exponent and since p_0>q_0 at least the first term of the numerator will still have a positive exponent, so we have something of the form c_in^y
    Which clearly goes to infinity

    Now let's assume p_0<q_0

    Then we do the same thing but we multiply top and bottom by \frac{\frac{1}{n^{p_0}}}{\frac{1}{n^{p_0}}}

    So everything on the top cancels or has a negative exponent, and at least the first term of the denominator has a positive exponent so we have something of the form c_i\frac{1}{n^y} which clearly goes to zero


    I can spell it fully out like the first example if needed, but I think it should be clear
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