Evaluate the limit: $lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}$












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$$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$



I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.






calculus limits-without-lhopital







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    1












    $begingroup$


    $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$



    I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.






    calculus limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$



      I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.






      calculus limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$



      I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.






      calculus limits-without-lhopital




      calculus limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 14:51









      Clement C.

      50.9k33992




      50.9k33992










      asked Jan 26 at 14:33









      user638473user638473

      505




      505






















          3 Answers
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          $begingroup$

          $$frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
          = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}=1+frac{1}{(x+3)(sqrt{x+6}+3)}$$






          share|cite|improve this answer









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          • 1




            $begingroup$
            Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
            $endgroup$
            – Michael Seifert
            Jan 26 at 15:12



















          4












          $begingroup$

          An elementary approach: "recognize the derivative."



          You can rewrite, for $xnotin{-3,3}$,
          $$
          frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
          = 1+ frac{sqrt{x+6}-3}{x^{2}-9}
          = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}
          = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)} tag{1}
          $$
          so the question boils down to computing $lim_{xto 3} frac{sqrt{x+6}-3}{x-3}$ (the rest is "under control").



          Now, we can recognize a derivative here: let $fcolon[0,infty) to [0,infty)$ be defined by
          $$
          f(x) = sqrt{x+6}, qquad xgeq 0,. tag{2}
          $$
          Note that $f$ is differentiable, with $f'(x) = frac{1}{2sqrt{x+6}}$. Therefore,
          $$
          frac{1}{6} = frac{1}{2sqrt{9}} = f'(3) = lim_{xto 3} frac{f(x)-f(3)}{x-3} = lim_{xto 3} frac{sqrt{x+6}-3}{x-3} tag{3}
          $$
          giving the desired limit.



          Putting it together, from (1) and (3) we get
          $$
          begin{align*}
          lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
          &= 1+ lim_{xto 3} frac{sqrt{x+6}-3}{(x-3)(x+3)}= 1+frac{1}{6}lim_{xto 3} frac{sqrt{x+6}-3}{x-3}
          \&= 1+frac{1}{36} = boxed{frac{37}{36}}
          end{align*}$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$
            $$ = lim_{xto 3} frac{x^{2}+sqrt{x+6}-9-3}{x^{2}-9}$$
            $$ = lim_{xto 3} frac{x^{2}-9+sqrt{x+6}-3}{x^{2}-9}$$
            $$ = lim_{xto 3} Big(frac{x^{2}-9}{x^{2}-9}+frac{sqrt{x+6}-3}{x^{2}-9}Big) $$
            $$ = lim_{xto 3} frac{x^{2}-9}{x^{2}-9} + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
            $$ = lim_{xto 3} 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
            $$ = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$



            After rationalizing the numerator to remove the radical:
            $$lim_{xto 3}frac{sqrt{x+6}-3}{x^{2}-9} cdot frac{sqrt{x+6}+3}{sqrt{x+6}+3} = lim_{xto 3}frac{x+6-9}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} $$



            After factoring $ x^2 - 9 $, simplifying, and direct substitution:
            $$ lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x-3)(x+3)(sqrt{x+6}+3)} = lim_{xto 3}frac{1}{(x+3)(sqrt{x+6}+3)} = frac{1}{6cdot 6} = frac{1}{36}$$



            Thus $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9} = 1 + frac{1}{36} = frac{36}{36} + frac{1}{36} $$
            $$= frac{37}{36}$$






            share|cite|improve this answer











            $endgroup$













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              3 Answers
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              4












              $begingroup$

              $$frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}=1+frac{1}{(x+3)(sqrt{x+6}+3)}$$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
                $endgroup$
                – Michael Seifert
                Jan 26 at 15:12
















              4












              $begingroup$

              $$frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}=1+frac{1}{(x+3)(sqrt{x+6}+3)}$$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
                $endgroup$
                – Michael Seifert
                Jan 26 at 15:12














              4












              4








              4





              $begingroup$

              $$frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}=1+frac{1}{(x+3)(sqrt{x+6}+3)}$$






              share|cite|improve this answer









              $endgroup$



              $$frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}=1+frac{1}{(x+3)(sqrt{x+6}+3)}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 26 at 14:55









              Aleksas DomarkasAleksas Domarkas

              1,54817




              1,54817








              • 1




                $begingroup$
                Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
                $endgroup$
                – Michael Seifert
                Jan 26 at 15:12














              • 1




                $begingroup$
                Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
                $endgroup$
                – Michael Seifert
                Jan 26 at 15:12








              1




              1




              $begingroup$
              Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
              $endgroup$
              – Michael Seifert
              Jan 26 at 15:12




              $begingroup$
              Just in case it's not clear, this relies on the fact that $(sqrt{x+6} - 3)(sqrt{x+6} + 3) = (x+6) - 3^2 = x - 3$.
              $endgroup$
              – Michael Seifert
              Jan 26 at 15:12











              4












              $begingroup$

              An elementary approach: "recognize the derivative."



              You can rewrite, for $xnotin{-3,3}$,
              $$
              frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{x^{2}-9}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}
              = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)} tag{1}
              $$
              so the question boils down to computing $lim_{xto 3} frac{sqrt{x+6}-3}{x-3}$ (the rest is "under control").



              Now, we can recognize a derivative here: let $fcolon[0,infty) to [0,infty)$ be defined by
              $$
              f(x) = sqrt{x+6}, qquad xgeq 0,. tag{2}
              $$
              Note that $f$ is differentiable, with $f'(x) = frac{1}{2sqrt{x+6}}$. Therefore,
              $$
              frac{1}{6} = frac{1}{2sqrt{9}} = f'(3) = lim_{xto 3} frac{f(x)-f(3)}{x-3} = lim_{xto 3} frac{sqrt{x+6}-3}{x-3} tag{3}
              $$
              giving the desired limit.



              Putting it together, from (1) and (3) we get
              $$
              begin{align*}
              lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
              &= 1+ lim_{xto 3} frac{sqrt{x+6}-3}{(x-3)(x+3)}= 1+frac{1}{6}lim_{xto 3} frac{sqrt{x+6}-3}{x-3}
              \&= 1+frac{1}{36} = boxed{frac{37}{36}}
              end{align*}$$






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                An elementary approach: "recognize the derivative."



                You can rewrite, for $xnotin{-3,3}$,
                $$
                frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                = 1+ frac{sqrt{x+6}-3}{x^{2}-9}
                = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}
                = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)} tag{1}
                $$
                so the question boils down to computing $lim_{xto 3} frac{sqrt{x+6}-3}{x-3}$ (the rest is "under control").



                Now, we can recognize a derivative here: let $fcolon[0,infty) to [0,infty)$ be defined by
                $$
                f(x) = sqrt{x+6}, qquad xgeq 0,. tag{2}
                $$
                Note that $f$ is differentiable, with $f'(x) = frac{1}{2sqrt{x+6}}$. Therefore,
                $$
                frac{1}{6} = frac{1}{2sqrt{9}} = f'(3) = lim_{xto 3} frac{f(x)-f(3)}{x-3} = lim_{xto 3} frac{sqrt{x+6}-3}{x-3} tag{3}
                $$
                giving the desired limit.



                Putting it together, from (1) and (3) we get
                $$
                begin{align*}
                lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                &= 1+ lim_{xto 3} frac{sqrt{x+6}-3}{(x-3)(x+3)}= 1+frac{1}{6}lim_{xto 3} frac{sqrt{x+6}-3}{x-3}
                \&= 1+frac{1}{36} = boxed{frac{37}{36}}
                end{align*}$$






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  An elementary approach: "recognize the derivative."



                  You can rewrite, for $xnotin{-3,3}$,
                  $$
                  frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                  = 1+ frac{sqrt{x+6}-3}{x^{2}-9}
                  = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}
                  = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)} tag{1}
                  $$
                  so the question boils down to computing $lim_{xto 3} frac{sqrt{x+6}-3}{x-3}$ (the rest is "under control").



                  Now, we can recognize a derivative here: let $fcolon[0,infty) to [0,infty)$ be defined by
                  $$
                  f(x) = sqrt{x+6}, qquad xgeq 0,. tag{2}
                  $$
                  Note that $f$ is differentiable, with $f'(x) = frac{1}{2sqrt{x+6}}$. Therefore,
                  $$
                  frac{1}{6} = frac{1}{2sqrt{9}} = f'(3) = lim_{xto 3} frac{f(x)-f(3)}{x-3} = lim_{xto 3} frac{sqrt{x+6}-3}{x-3} tag{3}
                  $$
                  giving the desired limit.



                  Putting it together, from (1) and (3) we get
                  $$
                  begin{align*}
                  lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                  &= 1+ lim_{xto 3} frac{sqrt{x+6}-3}{(x-3)(x+3)}= 1+frac{1}{6}lim_{xto 3} frac{sqrt{x+6}-3}{x-3}
                  \&= 1+frac{1}{36} = boxed{frac{37}{36}}
                  end{align*}$$






                  share|cite|improve this answer











                  $endgroup$



                  An elementary approach: "recognize the derivative."



                  You can rewrite, for $xnotin{-3,3}$,
                  $$
                  frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                  = 1+ frac{sqrt{x+6}-3}{x^{2}-9}
                  = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)}
                  = 1+ frac{sqrt{x+6}-3}{(x-3)(x+3)} tag{1}
                  $$
                  so the question boils down to computing $lim_{xto 3} frac{sqrt{x+6}-3}{x-3}$ (the rest is "under control").



                  Now, we can recognize a derivative here: let $fcolon[0,infty) to [0,infty)$ be defined by
                  $$
                  f(x) = sqrt{x+6}, qquad xgeq 0,. tag{2}
                  $$
                  Note that $f$ is differentiable, with $f'(x) = frac{1}{2sqrt{x+6}}$. Therefore,
                  $$
                  frac{1}{6} = frac{1}{2sqrt{9}} = f'(3) = lim_{xto 3} frac{f(x)-f(3)}{x-3} = lim_{xto 3} frac{sqrt{x+6}-3}{x-3} tag{3}
                  $$
                  giving the desired limit.



                  Putting it together, from (1) and (3) we get
                  $$
                  begin{align*}
                  lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9}
                  &= 1+ lim_{xto 3} frac{sqrt{x+6}-3}{(x-3)(x+3)}= 1+frac{1}{6}lim_{xto 3} frac{sqrt{x+6}-3}{x-3}
                  \&= 1+frac{1}{36} = boxed{frac{37}{36}}
                  end{align*}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 26 at 14:51

























                  answered Jan 26 at 14:45









                  Clement C.Clement C.

                  50.9k33992




                  50.9k33992























                      1












                      $begingroup$

                      $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$
                      $$ = lim_{xto 3} frac{x^{2}+sqrt{x+6}-9-3}{x^{2}-9}$$
                      $$ = lim_{xto 3} frac{x^{2}-9+sqrt{x+6}-3}{x^{2}-9}$$
                      $$ = lim_{xto 3} Big(frac{x^{2}-9}{x^{2}-9}+frac{sqrt{x+6}-3}{x^{2}-9}Big) $$
                      $$ = lim_{xto 3} frac{x^{2}-9}{x^{2}-9} + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                      $$ = lim_{xto 3} 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                      $$ = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$



                      After rationalizing the numerator to remove the radical:
                      $$lim_{xto 3}frac{sqrt{x+6}-3}{x^{2}-9} cdot frac{sqrt{x+6}+3}{sqrt{x+6}+3} = lim_{xto 3}frac{x+6-9}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} $$



                      After factoring $ x^2 - 9 $, simplifying, and direct substitution:
                      $$ lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x-3)(x+3)(sqrt{x+6}+3)} = lim_{xto 3}frac{1}{(x+3)(sqrt{x+6}+3)} = frac{1}{6cdot 6} = frac{1}{36}$$



                      Thus $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9} = 1 + frac{1}{36} = frac{36}{36} + frac{1}{36} $$
                      $$= frac{37}{36}$$






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$
                        $$ = lim_{xto 3} frac{x^{2}+sqrt{x+6}-9-3}{x^{2}-9}$$
                        $$ = lim_{xto 3} frac{x^{2}-9+sqrt{x+6}-3}{x^{2}-9}$$
                        $$ = lim_{xto 3} Big(frac{x^{2}-9}{x^{2}-9}+frac{sqrt{x+6}-3}{x^{2}-9}Big) $$
                        $$ = lim_{xto 3} frac{x^{2}-9}{x^{2}-9} + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                        $$ = lim_{xto 3} 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                        $$ = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$



                        After rationalizing the numerator to remove the radical:
                        $$lim_{xto 3}frac{sqrt{x+6}-3}{x^{2}-9} cdot frac{sqrt{x+6}+3}{sqrt{x+6}+3} = lim_{xto 3}frac{x+6-9}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} $$



                        After factoring $ x^2 - 9 $, simplifying, and direct substitution:
                        $$ lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x-3)(x+3)(sqrt{x+6}+3)} = lim_{xto 3}frac{1}{(x+3)(sqrt{x+6}+3)} = frac{1}{6cdot 6} = frac{1}{36}$$



                        Thus $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9} = 1 + frac{1}{36} = frac{36}{36} + frac{1}{36} $$
                        $$= frac{37}{36}$$






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$
                          $$ = lim_{xto 3} frac{x^{2}+sqrt{x+6}-9-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} frac{x^{2}-9+sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} Big(frac{x^{2}-9}{x^{2}-9}+frac{sqrt{x+6}-3}{x^{2}-9}Big) $$
                          $$ = lim_{xto 3} frac{x^{2}-9}{x^{2}-9} + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$



                          After rationalizing the numerator to remove the radical:
                          $$lim_{xto 3}frac{sqrt{x+6}-3}{x^{2}-9} cdot frac{sqrt{x+6}+3}{sqrt{x+6}+3} = lim_{xto 3}frac{x+6-9}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} $$



                          After factoring $ x^2 - 9 $, simplifying, and direct substitution:
                          $$ lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x-3)(x+3)(sqrt{x+6}+3)} = lim_{xto 3}frac{1}{(x+3)(sqrt{x+6}+3)} = frac{1}{6cdot 6} = frac{1}{36}$$



                          Thus $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9} = 1 + frac{1}{36} = frac{36}{36} + frac{1}{36} $$
                          $$= frac{37}{36}$$






                          share|cite|improve this answer











                          $endgroup$



                          $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} $$
                          $$ = lim_{xto 3} frac{x^{2}+sqrt{x+6}-9-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} frac{x^{2}-9+sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} Big(frac{x^{2}-9}{x^{2}-9}+frac{sqrt{x+6}-3}{x^{2}-9}Big) $$
                          $$ = lim_{xto 3} frac{x^{2}-9}{x^{2}-9} + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = lim_{xto 3} 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$
                          $$ = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9}$$



                          After rationalizing the numerator to remove the radical:
                          $$lim_{xto 3}frac{sqrt{x+6}-3}{x^{2}-9} cdot frac{sqrt{x+6}+3}{sqrt{x+6}+3} = lim_{xto 3}frac{x+6-9}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} $$



                          After factoring $ x^2 - 9 $, simplifying, and direct substitution:
                          $$ lim_{xto 3}frac{x-3}{(x^{2}-9)(sqrt{x+6}+3)} = lim_{xto 3}frac{x-3}{(x-3)(x+3)(sqrt{x+6}+3)} = lim_{xto 3}frac{1}{(x+3)(sqrt{x+6}+3)} = frac{1}{6cdot 6} = frac{1}{36}$$



                          Thus $$lim_{xto 3} frac{x^{2}+sqrt{x+6}-12}{x^{2}-9} = 1 + lim_{xto 3} frac{sqrt{x+6}-3}{x^{2}-9} = 1 + frac{1}{36} = frac{36}{36} + frac{1}{36} $$
                          $$= frac{37}{36}$$







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                          edited Jan 27 at 21:06

























                          answered Jan 26 at 16:30









                          Marvin CohenMarvin Cohen

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