Solve the limit $lim_{xto2}({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2})}$












0












$begingroup$


$$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



What I tried: Got the fraction to the same denominator
$$begin{align}
lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
&=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
&={frac{-4+6-1}{0}}
end{align}$$



There is a mistake that I can't find...






calculus limits







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$endgroup$

















    0












    $begingroup$


    $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



    What I tried: Got the fraction to the same denominator
    $$begin{align}
    lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
    &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
    &={frac{-4+6-1}{0}}
    end{align}$$



    There is a mistake that I can't find...






    calculus limits







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



      What I tried: Got the fraction to the same denominator
      $$begin{align}
      lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
      &={frac{-4+6-1}{0}}
      end{align}$$



      There is a mistake that I can't find...






      calculus limits







      share|cite|improve this question











      $endgroup$




      $$lim_{xto2}left(frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}right)$$



      What I tried: Got the fraction to the same denominator
      $$begin{align}
      lim_{xto2}{frac{x^2-3x+2-x(x-2)^2}{x(x-2)^2(x^2-3x+2)}}&=lim_{xto2}{frac{(x-2)(x-1)-x(x-2)^2}{x(x-2)^2(x-2)(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{x-1-x^2+2x}{x(x-2)^2(x-1)}}\
      &=lim_{xto2}{frac{-x^2+3x-1}{x(x-2)^2(x-1)}}\
      &={frac{-4+6-1}{0}}
      end{align}$$



      There is a mistake that I can't find...






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 14:50









      mrtaurho

      4,23421234




      4,23421234










      asked Dec 8 '18 at 14:33









      Bili DebiliBili Debili

      1428




      1428






















          3 Answers
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          $begingroup$

          The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



          enter image description here






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              By $y=x-2 to 0$ we have that



              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



              and



              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                $endgroup$
                – gimusi
                Dec 8 '18 at 18:58











              Your Answer





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              3 Answers
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              active

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              3 Answers
              3






              active

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              active

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              active

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              3












              $begingroup$

              The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  The numerator tend to a positive number, the denominator goes to $0$ from the positive directione. The sequence actually goes to infinity.



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 14:40









                  Siong Thye GohSiong Thye Goh

                  101k1466117




                  101k1466117























                      2












                      $begingroup$

                      Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$






                          share|cite|improve this answer









                          $endgroup$



                          Hint: Use that $$x^2-3x+2-x(x-2)^2=-(x-2)(x^2-3x+1)$$ and $$x^2-3x+2=(x-1)(x-2)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 8 '18 at 14:39









                          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                          74.6k42865




                          74.6k42865























                              1












                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58
















                              1












                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58














                              1












                              1








                              1





                              $begingroup$

                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$






                              share|cite|improve this answer











                              $endgroup$



                              By $y=x-2 to 0$ we have that



                              $$lim_{xto 2} left({frac{1}{x(x-2)^2}-frac{1}{x^2-3x+2}}right)=lim_{yto 0} left({frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}right)$$



                              and



                              $${frac{1}{y^2(y+2)}-frac{1}{y(y+1)}}={frac{y+1-y^2-2y}{y^2(y+1)(y+2)}}=frac1ycdot{frac{-y^2-y+1}{(y+1)(y+2)}}to infty$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 8 '18 at 18:59

























                              answered Dec 8 '18 at 14:41









                              gimusigimusi

                              92.8k84494




                              92.8k84494












                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58


















                              • $begingroup$
                                @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 18:58
















                              $begingroup$
                              @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                              $endgroup$
                              – gimusi
                              Dec 8 '18 at 18:58




                              $begingroup$
                              @user376343 mmmmhhhhh yes I thing there is something wrong with that! I fix...thanks
                              $endgroup$
                              – gimusi
                              Dec 8 '18 at 18:58


















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