Finding function given limit











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$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?










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    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    – Rebellos
    5 hours ago








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    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    – JavaMan
    5 hours ago















up vote
3
down vote

favorite












$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?










share|cite|improve this question









New contributor




Amanda Choi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    – Rebellos
    5 hours ago








  • 1




    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    – JavaMan
    5 hours ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?










share|cite|improve this question









New contributor




Amanda Choi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?







limits






share|cite|improve this question









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Amanda Choi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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share|cite|improve this question




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edited 4 hours ago









gimusi

86.6k74393




86.6k74393






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asked 5 hours ago









Amanda Choi

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161




New contributor




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Check out our Code of Conduct.





New contributor





Amanda Choi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Amanda Choi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    – Rebellos
    5 hours ago








  • 1




    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    – JavaMan
    5 hours ago














  • 1




    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    – Rebellos
    5 hours ago








  • 1




    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    – JavaMan
    5 hours ago








1




1




Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
– Rebellos
5 hours ago






Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
– Rebellos
5 hours ago






1




1




Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
– JavaMan
5 hours ago




Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
– JavaMan
5 hours ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote













Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



$frac{2+a}4=3$ ---> $a=10$



Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



Then $c=-8, d=-20$.






share|cite|improve this answer






























    up vote
    1
    down vote













    We have that



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



    requires that for $x=2$



    $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



    then



    $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



    then



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



    As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



    from which we can find $c$ and then $d$.






    share|cite|improve this answer























    • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
      – Christopher Marley
      4 hours ago






    • 1




      @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
      – gimusi
      4 hours ago











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    up vote
    3
    down vote













    Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



    Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



    $frac{2+a}4=3$ ---> $a=10$



    Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



    Then $c=-8, d=-20$.






    share|cite|improve this answer



























      up vote
      3
      down vote













      Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



      Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



      $frac{2+a}4=3$ ---> $a=10$



      Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



      Then $c=-8, d=-20$.






      share|cite|improve this answer

























        up vote
        3
        down vote










        up vote
        3
        down vote









        Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



        Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



        $frac{2+a}4=3$ ---> $a=10$



        Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



        Then $c=-8, d=-20$.






        share|cite|improve this answer














        Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



        Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



        $frac{2+a}4=3$ ---> $a=10$



        Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



        Then $c=-8, d=-20$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 4 hours ago









        Christopher Marley

        85814




        85814






















            up vote
            1
            down vote













            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer























            • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              – Christopher Marley
              4 hours ago






            • 1




              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              – gimusi
              4 hours ago















            up vote
            1
            down vote













            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer























            • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              – Christopher Marley
              4 hours ago






            • 1




              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              – gimusi
              4 hours ago













            up vote
            1
            down vote










            up vote
            1
            down vote









            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer














            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            gimusi

            86.6k74393




            86.6k74393












            • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              – Christopher Marley
              4 hours ago






            • 1




              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              – gimusi
              4 hours ago


















            • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              – Christopher Marley
              4 hours ago






            • 1




              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              – gimusi
              4 hours ago
















            Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
            – Christopher Marley
            4 hours ago




            Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
            – Christopher Marley
            4 hours ago




            1




            1




            @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
            – gimusi
            4 hours ago




            @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
            – gimusi
            4 hours ago










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