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Thread: limit 2

  1. #1
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    limit 2

    http://mustafayagci.com.tr/PDF/MY_M3.pd

    Page 156 question 30.
    Find a.b

    I tried to get x^2 out of the parantheses and tried to use L'hospital rule.
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  2. #2
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    Re: limit 2

    Quote Originally Posted by kastamonu View Post
    http://mustafayagci.com.tr/PDF/MY_M3.pdf

    Page 156 question 30.
    Find a.b

    I tried to get x^2 out of the parantheses and tried to use L'hospital rule.
    fixed link ...
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  3. #3
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    Re: limit 2

    $\begin{align*}\displaystyle \lim_{x \to \infty} \left( \dfrac{x^2}{x^2+1} \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}} + (a-3)x+b-1\right) & =5 \\ \left(\displaystyle \lim_{x \to \infty} \dfrac{1}{1+\cancel{\dfrac{1}{x^2}}^0} + \lim_{x\to \infty}(a-3)x\right) +b-1 & = 5 \\ 1+\lim_{x\to\infty}(a-3)x + b - 1 & = 5 \end{align*}$
    Last edited by SlipEternal; May 19th 2017 at 12:03 PM.
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    Re: limit 2

    This is where I left. If I take x out of parantheses I can't find a. I have only one choice. I will take a=3. And b=5.
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  5. #5
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    Re: limit 2

    Quote Originally Posted by kastamonu View Post
    This is where I left. If I take x out of parantheses I can't find a. I have only one choice. I will take a=3. And b=5.
    That is correct: $3\cdot 5=15.\;\;\; \bf{A}$
    Last edited by Plato; May 19th 2017 at 01:43 PM.
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    Re: limit 2

    \lim_{x\to \infty}px for p anything other than 0 is "infinity" (or "does not exist"). In order that this have any finite limit, a- 3 must be 0 so a must be 3. In that case, the limit is 1- b- 1= 5 so that b= 5.
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  7. #7
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    Re: limit 2

    Many Thanks.
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