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Thread: Limit problem (I made it up and worked it out.)

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    Limit problem (I made it up and worked it out.)

    The following limit exists. Find the value of the limit.


    $\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
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    Re: Limit problem (I made it up and worked it out.)

    Hey greg1313.

    Can you please show us what you have tried?

    What limit laws have you studied?
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    Re: Limit problem (I made it up and worked it out.)

    I would start by rationalising the numerator...
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by chiro View Post
    Hey greg1313.

    Can you please show us what you have tried?

    What limit laws have you studied?
    It's a math challenge problem. I'm challenging the rest of you.

    And notice that I stated that I worked it out.
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    Re: Limit problem (I made it up and worked it out.)

    \tfrac16
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by Archie View Post
    \tfrac16
    You're going to have to show on here how you got that alleged answer or at least an outline of the work.
    For instance, I could just graph f(x) = the expression that the limit is being taken and estimate the function
    value where the "hole" is at where x = 16.
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    Re: Limit problem (I made it up and worked it out.)

    (Edit)

    The following limit exists. Find the value of the limit. Please show your work.


    $\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by greg1313 View Post
    (Edit)

    The following limit exists. Find the value of the limit. Please show your work.


    $\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
    \[\frac{{(\sqrt x - 4)(\sqrt {x - 7\;} - 3)}}{{(\sqrt {x - 12\;} - 2)(\sqrt {x - 15\;} - 1)}} = \frac{{(\sqrt {x - 12\;} + 2)(\sqrt {x - 15\;} + 1)}}{{(\sqrt x + 4)(\sqrt {x - 7\;} + 3)}}\]
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by greg1313 View Post
    Please show your work.
    LOL
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    Re: Limit problem (I made it up and worked it out.)

    Multiply top and bottom by the conjugates of the four terms. The result is Plato's expression.

    I posted no working so as not to spoil it for anyone else.
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    Re: Limit problem (I made it up and worked it out.)

    Nice algebra problem. Thanks.

    Prove It was on the right track for a formal demonstration of Archie's result. And of course Plato's answer is elegant. For anyone who wants detail, see below:

    Spoiler:
    $x \ne 16 \implies \dfrac{\sqrt{x} - 4}{\sqrt{x - 12} - 2} =\dfrac{\sqrt{x} - 4}{\sqrt{x - 12} - 2} * \dfrac{\sqrt{x} +4}{\sqrt{x} + 4} = \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)}$.

    $x \ne 16 \implies \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)} = \dfrac{x - 16}{(\sqrt{x - 12} - 2)(\sqrt{x} + 4)} * \dfrac{\sqrt{x - 12} + 2}{\sqrt{x - 12} + 2} =$

    $\dfrac{(x - 16)(\sqrt{x - 12} + 2)}{(\{x - 12\} - 4)(\sqrt{x} + 4)} = \dfrac{\cancel {(x - 16) }(\sqrt{x - 12} + 2)}{\cancel {(x - 16)}(\sqrt{x} + 4)} = \dfrac{\sqrt{x - 12} + 2}{\sqrt{x} + 4)}.$

    $\displaystyle \therefore \lim_{x \rightarrow 16}\dfrac{\sqrt{x - 12} + 2}{\sqrt{x} + 4)} = \dfrac{\sqrt{16 - 12} + 2 }{\sqrt{16} + 4} = \dfrac{2 + 2}{4 + 4} = \dfrac{1}{2}.$

    $x \ne 16 \implies \dfrac{\sqrt{x - 7} - 3}{\sqrt{x - 15} - 1} = \dfrac{\sqrt{x - 7} - 3}{\sqrt{x - 15} - 1} * \dfrac{\sqrt{x - 7} + 3}{\sqrt{x - 7} + 3} = \dfrac{x - 7 - 9}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} = \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)}.$

    $x \ne 16 \implies \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} = \dfrac{x - 16}{(\sqrt{x - 15} - 1)(\sqrt{x - 7} + 3)} * \dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 15} + 1} =$

    $\dfrac{(x - 16)(\sqrt{x - 15} + 1)}{(\{x - 15\} - 1)(\sqrt{x - 7} + 3)} = \dfrac{\cancel {(x - 16)}(\sqrt{x - 15} + 1)}{\cancel {(x - 16)}(\sqrt{x - 7} + 3)} = \dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 7} + 3}.$

    $\displaystyle \therefore \lim_{x \rightarrow 16}\dfrac{\sqrt{x - 15} + 1}{\sqrt{x - 7} + 3} = \dfrac{\sqrt{16 - 1} + 1}{\sqrt{16 - 7} + 3} = \dfrac{1 + 1}{3 + 3} = \dfrac{1}{3}.$

    $THUS, \displaystyle \lim_{x \rightarrow 16} \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7} - 3)}{(\sqrt{x - 12} - 2)(\sqrt{x - 15} - 1)} = \dfrac{1}{2} * \dfrac{1}{3} = \dfrac{1}{6}.$
    Last edited by JeffM; Jan 26th 2017 at 09:42 AM.
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by Archie View Post
    > > > Multiply top and bottom by the conjugates of the four terms. < < < The result is Plato's expression.
    Archie, this is at least an outline of what to do. JeffM, Plato's post is not elegant. It is a relatively compact form step in
    the process after doing what you stated. Someone has to understand/justify how to get there in the first place.
    Last edited by greg1313; Jan 26th 2017 at 11:24 AM.
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by greg1313 View Post
    JeffM, Plato's post is not elegant... Someone has to understand/justify how to get there in the first place.
    Tastes differ. In my opinion, his statement is an elegant consequence of utilizing conjugates and indicates why it is a neat problem in the first place. I do agree that a beginning student is highly unlikely to follow so someone should provide a detailed development (as I did). But that does not detract from its simplicity and clarity (for those familiar with conjugates).
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    Re: Limit problem (I made it up and worked it out.)

    In terms of limits solved using conjugate pairs, I prefer limits of the form
    \lim_{x \to \infty} \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)
    because the convergence is, to me, somewhat counter-intuitive.

    If memory serves, I recently saw an example that combined the two ideas.
    Last edited by Archie; Jan 26th 2017 at 03:05 PM.
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    Re: Limit problem (I made it up and worked it out.)

    Quote Originally Posted by greg1313 View Post
    Archie, this is at least an outline of what to do. JeffM, Plato's post is not elegant. It is a relatively compact form step in the process after doing what you stated. Someone has to understand/justify how to get there in the first place.
    Quote Originally Posted by Archie View Post
    In terms of limits solved using conjugate pairs, I prefer limits of the form
    \lim_{x \to \infty} \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)
    because the convergence is, to me, somewhat counter-intuitive.
    For those of us who have had a life-time of experience teaching/writing about this topic surely realize that students working on these types of limits should have a solid grounding in basic algebra manipulation of radicals. Any of us should have taught the time honored 'trick' of inverse/sign-change. That is all that I posted. It skipped no steps, it left nothing out that someone competent is basic algebra might fail to see.

    Here is that trick again.
    \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)=\dfrac{(a-c)x+(b-d)}{\sqrt{x^2 + ax + b} + \sqrt{x^2 + cx + d}}\to\dfrac{a-c}{\sqrt{2}} as $x\to\infty$.
    Last edited by Plato; Jan 26th 2017 at 06:36 PM.
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