Limit to infinity of polynomial fraction












0












$begingroup$


I'm trying to solve for the function this limit approaches:



$$lim_{xto infty} frac{x^2}{x-2}$$



From graphing the function online I know it approaches $y=x+2$, but I am just unsure how to go about proving it. I can understand why it approaches $y=x$ but I'm unsure as to where the $+2$ comes from.



Any help would be greatly appreciated






limits polynomials graphing-functions







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  • 2




    $begingroup$
    $frac{x^2}{x-2}=x+frac{4}{x-2}+2$
    $endgroup$
    – coreyman317
    Feb 3 at 2:08
















0












$begingroup$


I'm trying to solve for the function this limit approaches:



$$lim_{xto infty} frac{x^2}{x-2}$$



From graphing the function online I know it approaches $y=x+2$, but I am just unsure how to go about proving it. I can understand why it approaches $y=x$ but I'm unsure as to where the $+2$ comes from.



Any help would be greatly appreciated






limits polynomials graphing-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $frac{x^2}{x-2}=x+frac{4}{x-2}+2$
    $endgroup$
    – coreyman317
    Feb 3 at 2:08














0












0








0





$begingroup$


I'm trying to solve for the function this limit approaches:



$$lim_{xto infty} frac{x^2}{x-2}$$



From graphing the function online I know it approaches $y=x+2$, but I am just unsure how to go about proving it. I can understand why it approaches $y=x$ but I'm unsure as to where the $+2$ comes from.



Any help would be greatly appreciated






limits polynomials graphing-functions







share|cite|improve this question









$endgroup$




I'm trying to solve for the function this limit approaches:



$$lim_{xto infty} frac{x^2}{x-2}$$



From graphing the function online I know it approaches $y=x+2$, but I am just unsure how to go about proving it. I can understand why it approaches $y=x$ but I'm unsure as to where the $+2$ comes from.



Any help would be greatly appreciated






limits polynomials graphing-functions




limits polynomials graphing-functions






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asked Feb 3 at 2:01









ultralightultralight

806




806








  • 2




    $begingroup$
    $frac{x^2}{x-2}=x+frac{4}{x-2}+2$
    $endgroup$
    – coreyman317
    Feb 3 at 2:08














  • 2




    $begingroup$
    $frac{x^2}{x-2}=x+frac{4}{x-2}+2$
    $endgroup$
    – coreyman317
    Feb 3 at 2:08








2




2




$begingroup$
$frac{x^2}{x-2}=x+frac{4}{x-2}+2$
$endgroup$
– coreyman317
Feb 3 at 2:08




$begingroup$
$frac{x^2}{x-2}=x+frac{4}{x-2}+2$
$endgroup$
– coreyman317
Feb 3 at 2:08










1 Answer
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$begingroup$

Let us prove your statement by simply showing $$frac{x^2}{x-2}=x+frac{4}{x-2}+2$$



So $$frac{x^2}{x-2}=frac{x^2-4+4}{x-2}=frac{x^2-4}{x-2}+frac{4}{x-2}=frac{(x-2)(x+2)}{x-2}+frac{4}{x-2}=x+2+frac{4}{x-2}$$



Now take the limit as $xtoinfty$ of both sides: $$lim_{xtoinfty}frac{x^2}{x-2}=lim_{xtoinfty}left(x+2+frac{4}{x-2}right)=lim_{xtoinfty}left(x+2right)+lim_{xtoinfty}frac{4}{x-2}=lim_{xtoinfty}left(x+2right)$$






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    1 Answer
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    3












    $begingroup$

    Let us prove your statement by simply showing $$frac{x^2}{x-2}=x+frac{4}{x-2}+2$$



    So $$frac{x^2}{x-2}=frac{x^2-4+4}{x-2}=frac{x^2-4}{x-2}+frac{4}{x-2}=frac{(x-2)(x+2)}{x-2}+frac{4}{x-2}=x+2+frac{4}{x-2}$$



    Now take the limit as $xtoinfty$ of both sides: $$lim_{xtoinfty}frac{x^2}{x-2}=lim_{xtoinfty}left(x+2+frac{4}{x-2}right)=lim_{xtoinfty}left(x+2right)+lim_{xtoinfty}frac{4}{x-2}=lim_{xtoinfty}left(x+2right)$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Let us prove your statement by simply showing $$frac{x^2}{x-2}=x+frac{4}{x-2}+2$$



      So $$frac{x^2}{x-2}=frac{x^2-4+4}{x-2}=frac{x^2-4}{x-2}+frac{4}{x-2}=frac{(x-2)(x+2)}{x-2}+frac{4}{x-2}=x+2+frac{4}{x-2}$$



      Now take the limit as $xtoinfty$ of both sides: $$lim_{xtoinfty}frac{x^2}{x-2}=lim_{xtoinfty}left(x+2+frac{4}{x-2}right)=lim_{xtoinfty}left(x+2right)+lim_{xtoinfty}frac{4}{x-2}=lim_{xtoinfty}left(x+2right)$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Let us prove your statement by simply showing $$frac{x^2}{x-2}=x+frac{4}{x-2}+2$$



        So $$frac{x^2}{x-2}=frac{x^2-4+4}{x-2}=frac{x^2-4}{x-2}+frac{4}{x-2}=frac{(x-2)(x+2)}{x-2}+frac{4}{x-2}=x+2+frac{4}{x-2}$$



        Now take the limit as $xtoinfty$ of both sides: $$lim_{xtoinfty}frac{x^2}{x-2}=lim_{xtoinfty}left(x+2+frac{4}{x-2}right)=lim_{xtoinfty}left(x+2right)+lim_{xtoinfty}frac{4}{x-2}=lim_{xtoinfty}left(x+2right)$$






        share|cite|improve this answer









        $endgroup$



        Let us prove your statement by simply showing $$frac{x^2}{x-2}=x+frac{4}{x-2}+2$$



        So $$frac{x^2}{x-2}=frac{x^2-4+4}{x-2}=frac{x^2-4}{x-2}+frac{4}{x-2}=frac{(x-2)(x+2)}{x-2}+frac{4}{x-2}=x+2+frac{4}{x-2}$$



        Now take the limit as $xtoinfty$ of both sides: $$lim_{xtoinfty}frac{x^2}{x-2}=lim_{xtoinfty}left(x+2+frac{4}{x-2}right)=lim_{xtoinfty}left(x+2right)+lim_{xtoinfty}frac{4}{x-2}=lim_{xtoinfty}left(x+2right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 3 at 2:14









        coreyman317coreyman317

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