solve $lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}$











up vote
5
down vote

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I need to find
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}$$



Looking at the graph, I know the answer should be $frac{20}{17}$, but when I tried solving it, I reached $0$.



Here are the two ways I approached this:



WAY I:



$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{x^2}(2- frac{50}{x^2})} {require{cancel} cancel{x^2}(2+ frac{3}{x}-frac{35}{x^2})}
=frac{2-2}{frac {42}{5}}=0
$$



WAY II:
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{2}(x^2- 25)} {require{cancel} cancel{2}(x^2+ frac{3}{2}x-frac{35}{2})}
=lim_{xrightarrow -5} frac{{require{cancel} cancel{(x-5)}}(x+5)}{{require{cancel} cancel{(x-5)}}(x+3.5)}= frac{-5+5}{-5+3.5}=0
$$



What am I doing wrong here?



Thanks!






calculus limits







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  • Generally $0$ comes when we assume something or enforce partial limits.
    – Akash Roy
    Nov 22 at 17:03






  • 1




    On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
    – Love Invariants
    Nov 22 at 17:09










  • @LoveInvariants you mean on step 2 of way 2? the factorization?
    – Netanel
    Nov 22 at 17:17















up vote
5
down vote

favorite












I need to find
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}$$



Looking at the graph, I know the answer should be $frac{20}{17}$, but when I tried solving it, I reached $0$.



Here are the two ways I approached this:



WAY I:



$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{x^2}(2- frac{50}{x^2})} {require{cancel} cancel{x^2}(2+ frac{3}{x}-frac{35}{x^2})}
=frac{2-2}{frac {42}{5}}=0
$$



WAY II:
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{2}(x^2- 25)} {require{cancel} cancel{2}(x^2+ frac{3}{2}x-frac{35}{2})}
=lim_{xrightarrow -5} frac{{require{cancel} cancel{(x-5)}}(x+5)}{{require{cancel} cancel{(x-5)}}(x+3.5)}= frac{-5+5}{-5+3.5}=0
$$



What am I doing wrong here?



Thanks!






calculus limits







share|cite|improve this question
























  • Generally $0$ comes when we assume something or enforce partial limits.
    – Akash Roy
    Nov 22 at 17:03






  • 1




    On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
    – Love Invariants
    Nov 22 at 17:09










  • @LoveInvariants you mean on step 2 of way 2? the factorization?
    – Netanel
    Nov 22 at 17:17













up vote
5
down vote

favorite









up vote
5
down vote

favorite











I need to find
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}$$



Looking at the graph, I know the answer should be $frac{20}{17}$, but when I tried solving it, I reached $0$.



Here are the two ways I approached this:



WAY I:



$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{x^2}(2- frac{50}{x^2})} {require{cancel} cancel{x^2}(2+ frac{3}{x}-frac{35}{x^2})}
=frac{2-2}{frac {42}{5}}=0
$$



WAY II:
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{2}(x^2- 25)} {require{cancel} cancel{2}(x^2+ frac{3}{2}x-frac{35}{2})}
=lim_{xrightarrow -5} frac{{require{cancel} cancel{(x-5)}}(x+5)}{{require{cancel} cancel{(x-5)}}(x+3.5)}= frac{-5+5}{-5+3.5}=0
$$



What am I doing wrong here?



Thanks!






calculus limits







share|cite|improve this question















I need to find
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}$$



Looking at the graph, I know the answer should be $frac{20}{17}$, but when I tried solving it, I reached $0$.



Here are the two ways I approached this:



WAY I:



$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{x^2}(2- frac{50}{x^2})} {require{cancel} cancel{x^2}(2+ frac{3}{x}-frac{35}{x^2})}
=frac{2-2}{frac {42}{5}}=0
$$



WAY II:
$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35} =
lim_{xrightarrow -5} frac{require{cancel} cancel{2}(x^2- 25)} {require{cancel} cancel{2}(x^2+ frac{3}{2}x-frac{35}{2})}
=lim_{xrightarrow -5} frac{{require{cancel} cancel{(x-5)}}(x+5)}{{require{cancel} cancel{(x-5)}}(x+3.5)}= frac{-5+5}{-5+3.5}=0
$$



What am I doing wrong here?



Thanks!






calculus limits




calculus limits






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edited Nov 22 at 17:10

























asked Nov 22 at 16:59









Netanel

474




474












  • Generally $0$ comes when we assume something or enforce partial limits.
    – Akash Roy
    Nov 22 at 17:03






  • 1




    On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
    – Love Invariants
    Nov 22 at 17:09










  • @LoveInvariants you mean on step 2 of way 2? the factorization?
    – Netanel
    Nov 22 at 17:17


















  • Generally $0$ comes when we assume something or enforce partial limits.
    – Akash Roy
    Nov 22 at 17:03






  • 1




    On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
    – Love Invariants
    Nov 22 at 17:09










  • @LoveInvariants you mean on step 2 of way 2? the factorization?
    – Netanel
    Nov 22 at 17:17
















Generally $0$ comes when we assume something or enforce partial limits.
– Akash Roy
Nov 22 at 17:03




Generally $0$ comes when we assume something or enforce partial limits.
– Akash Roy
Nov 22 at 17:03




1




1




On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
– Love Invariants
Nov 22 at 17:09




On multiplying the denominators you will get $-3over2$ not $+3over2$ as coefficient of x.
– Love Invariants
Nov 22 at 17:09












@LoveInvariants you mean on step 2 of way 2? the factorization?
– Netanel
Nov 22 at 17:17




@LoveInvariants you mean on step 2 of way 2? the factorization?
– Netanel
Nov 22 at 17:17










4 Answers
4






active

oldest

votes

















up vote
6
down vote



accepted










For your way $1$, check the computation of your denominator, it should give you $0$ again.



For your way $2$, check your factorization in your denominator as well.



Use L'hopital's rule:



$$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}= lim_{xrightarrow -5} frac{4x}{4x+3}=frac{-20}{-17}=frac{20}{17}$$






share|cite|improve this answer





















  • Thank you. Can you point out the mistake in Way 2?
    – Netanel
    Nov 22 at 17:17






  • 3




    $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
    – Siong Thye Goh
    Nov 22 at 17:19




















up vote
7
down vote













Numerator



$2x^2-50=2(x-5)(x+5)$.



Denominator



$2x^2+3x -35 =(2x-7)(x+ 5)$



$dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$



$dfrac{2(x-5)}{2x+7}.$



Take the limit $x rightarrow -5.$



Try to factorize the original expression.The term $(x+5) $ cancels out .






share|cite|improve this answer





















  • Thank you. Why is my way of factorizing wrong though?
    – Netanel
    Nov 22 at 17:19






  • 1




    Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
    – Peter Szilas
    Nov 22 at 17:27


















up vote
5
down vote













Hint: Try factorization!



$$
frac{2x^2-50}{2x^2+3x-35}=frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
$$






share|cite|improve this answer




























    up vote
    5
    down vote













    As an alternative by $y=x+5 to 0$



    $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}=lim_{yrightarrow 0} frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=lim_{yrightarrow 0} frac{2y^2-20y}{2y^2-17y}=lim_{yrightarrow 0} frac{2y-20}{2y-17}$$






    share|cite|improve this answer





















    • Thank you. Why is my way wrong though?
      – Netanel
      Nov 22 at 17:18






    • 2




      @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
      – gimusi
      Nov 22 at 17:20






    • 1




      For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
      – gimusi
      Nov 22 at 17:40











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    For your way $1$, check the computation of your denominator, it should give you $0$ again.



    For your way $2$, check your factorization in your denominator as well.



    Use L'hopital's rule:



    $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}= lim_{xrightarrow -5} frac{4x}{4x+3}=frac{-20}{-17}=frac{20}{17}$$






    share|cite|improve this answer





















    • Thank you. Can you point out the mistake in Way 2?
      – Netanel
      Nov 22 at 17:17






    • 3




      $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
      – Siong Thye Goh
      Nov 22 at 17:19

















    up vote
    6
    down vote



    accepted










    For your way $1$, check the computation of your denominator, it should give you $0$ again.



    For your way $2$, check your factorization in your denominator as well.



    Use L'hopital's rule:



    $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}= lim_{xrightarrow -5} frac{4x}{4x+3}=frac{-20}{-17}=frac{20}{17}$$






    share|cite|improve this answer





















    • Thank you. Can you point out the mistake in Way 2?
      – Netanel
      Nov 22 at 17:17






    • 3




      $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
      – Siong Thye Goh
      Nov 22 at 17:19















    up vote
    6
    down vote



    accepted







    up vote
    6
    down vote



    accepted






    For your way $1$, check the computation of your denominator, it should give you $0$ again.



    For your way $2$, check your factorization in your denominator as well.



    Use L'hopital's rule:



    $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}= lim_{xrightarrow -5} frac{4x}{4x+3}=frac{-20}{-17}=frac{20}{17}$$






    share|cite|improve this answer












    For your way $1$, check the computation of your denominator, it should give you $0$ again.



    For your way $2$, check your factorization in your denominator as well.



    Use L'hopital's rule:



    $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}= lim_{xrightarrow -5} frac{4x}{4x+3}=frac{-20}{-17}=frac{20}{17}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 at 17:05









    Siong Thye Goh

    97.8k1463116




    97.8k1463116












    • Thank you. Can you point out the mistake in Way 2?
      – Netanel
      Nov 22 at 17:17






    • 3




      $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
      – Siong Thye Goh
      Nov 22 at 17:19




















    • Thank you. Can you point out the mistake in Way 2?
      – Netanel
      Nov 22 at 17:17






    • 3




      $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
      – Siong Thye Goh
      Nov 22 at 17:19


















    Thank you. Can you point out the mistake in Way 2?
    – Netanel
    Nov 22 at 17:17




    Thank you. Can you point out the mistake in Way 2?
    – Netanel
    Nov 22 at 17:17




    3




    3




    $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
    – Siong Thye Goh
    Nov 22 at 17:19






    $(x-5)(x+3.5)=x^2color{red}-1.5x -frac{35}2$. That is the factorization is not correct.
    – Siong Thye Goh
    Nov 22 at 17:19












    up vote
    7
    down vote













    Numerator



    $2x^2-50=2(x-5)(x+5)$.



    Denominator



    $2x^2+3x -35 =(2x-7)(x+ 5)$



    $dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$



    $dfrac{2(x-5)}{2x+7}.$



    Take the limit $x rightarrow -5.$



    Try to factorize the original expression.The term $(x+5) $ cancels out .






    share|cite|improve this answer





















    • Thank you. Why is my way of factorizing wrong though?
      – Netanel
      Nov 22 at 17:19






    • 1




      Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
      – Peter Szilas
      Nov 22 at 17:27















    up vote
    7
    down vote













    Numerator



    $2x^2-50=2(x-5)(x+5)$.



    Denominator



    $2x^2+3x -35 =(2x-7)(x+ 5)$



    $dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$



    $dfrac{2(x-5)}{2x+7}.$



    Take the limit $x rightarrow -5.$



    Try to factorize the original expression.The term $(x+5) $ cancels out .






    share|cite|improve this answer





















    • Thank you. Why is my way of factorizing wrong though?
      – Netanel
      Nov 22 at 17:19






    • 1




      Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
      – Peter Szilas
      Nov 22 at 17:27













    up vote
    7
    down vote










    up vote
    7
    down vote









    Numerator



    $2x^2-50=2(x-5)(x+5)$.



    Denominator



    $2x^2+3x -35 =(2x-7)(x+ 5)$



    $dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$



    $dfrac{2(x-5)}{2x+7}.$



    Take the limit $x rightarrow -5.$



    Try to factorize the original expression.The term $(x+5) $ cancels out .






    share|cite|improve this answer












    Numerator



    $2x^2-50=2(x-5)(x+5)$.



    Denominator



    $2x^2+3x -35 =(2x-7)(x+ 5)$



    $dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$



    $dfrac{2(x-5)}{2x+7}.$



    Take the limit $x rightarrow -5.$



    Try to factorize the original expression.The term $(x+5) $ cancels out .







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 at 17:18









    Peter Szilas

    10.6k2720




    10.6k2720












    • Thank you. Why is my way of factorizing wrong though?
      – Netanel
      Nov 22 at 17:19






    • 1




      Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
      – Peter Szilas
      Nov 22 at 17:27


















    • Thank you. Why is my way of factorizing wrong though?
      – Netanel
      Nov 22 at 17:19






    • 1




      Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
      – Peter Szilas
      Nov 22 at 17:27
















    Thank you. Why is my way of factorizing wrong though?
    – Netanel
    Nov 22 at 17:19




    Thank you. Why is my way of factorizing wrong though?
    – Netanel
    Nov 22 at 17:19




    1




    1




    Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
    – Peter Szilas
    Nov 22 at 17:27




    Netanel. The term (x+5) goes to zero,factor in numerator and denominator, cancels .I do not see it canceling in your calculation, so taking the limit numerator and denominator should go to zero .Please check .Denominator in Way1 =0!!
    – Peter Szilas
    Nov 22 at 17:27










    up vote
    5
    down vote













    Hint: Try factorization!



    $$
    frac{2x^2-50}{2x^2+3x-35}=frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
    $$






    share|cite|improve this answer

























      up vote
      5
      down vote













      Hint: Try factorization!



      $$
      frac{2x^2-50}{2x^2+3x-35}=frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
      $$






      share|cite|improve this answer























        up vote
        5
        down vote










        up vote
        5
        down vote









        Hint: Try factorization!



        $$
        frac{2x^2-50}{2x^2+3x-35}=frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
        $$






        share|cite|improve this answer












        Hint: Try factorization!



        $$
        frac{2x^2-50}{2x^2+3x-35}=frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 17:11









        Mefitico

        920117




        920117






















            up vote
            5
            down vote













            As an alternative by $y=x+5 to 0$



            $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}=lim_{yrightarrow 0} frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=lim_{yrightarrow 0} frac{2y^2-20y}{2y^2-17y}=lim_{yrightarrow 0} frac{2y-20}{2y-17}$$






            share|cite|improve this answer





















            • Thank you. Why is my way wrong though?
              – Netanel
              Nov 22 at 17:18






            • 2




              @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
              – gimusi
              Nov 22 at 17:20






            • 1




              For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
              – gimusi
              Nov 22 at 17:40















            up vote
            5
            down vote













            As an alternative by $y=x+5 to 0$



            $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}=lim_{yrightarrow 0} frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=lim_{yrightarrow 0} frac{2y^2-20y}{2y^2-17y}=lim_{yrightarrow 0} frac{2y-20}{2y-17}$$






            share|cite|improve this answer





















            • Thank you. Why is my way wrong though?
              – Netanel
              Nov 22 at 17:18






            • 2




              @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
              – gimusi
              Nov 22 at 17:20






            • 1




              For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
              – gimusi
              Nov 22 at 17:40













            up vote
            5
            down vote










            up vote
            5
            down vote









            As an alternative by $y=x+5 to 0$



            $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}=lim_{yrightarrow 0} frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=lim_{yrightarrow 0} frac{2y^2-20y}{2y^2-17y}=lim_{yrightarrow 0} frac{2y-20}{2y-17}$$






            share|cite|improve this answer












            As an alternative by $y=x+5 to 0$



            $$lim_{xrightarrow -5} frac{2x^2-50}{2x^2+3x-35}=lim_{yrightarrow 0} frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=lim_{yrightarrow 0} frac{2y^2-20y}{2y^2-17y}=lim_{yrightarrow 0} frac{2y-20}{2y-17}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 17:11









            gimusi

            92.7k94495




            92.7k94495












            • Thank you. Why is my way wrong though?
              – Netanel
              Nov 22 at 17:18






            • 2




              @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
              – gimusi
              Nov 22 at 17:20






            • 1




              For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
              – gimusi
              Nov 22 at 17:40


















            • Thank you. Why is my way wrong though?
              – Netanel
              Nov 22 at 17:18






            • 2




              @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
              – gimusi
              Nov 22 at 17:20






            • 1




              For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
              – gimusi
              Nov 22 at 17:40
















            Thank you. Why is my way wrong though?
            – Netanel
            Nov 22 at 17:18




            Thank you. Why is my way wrong though?
            – Netanel
            Nov 22 at 17:18




            2




            2




            @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
            – gimusi
            Nov 22 at 17:20




            @Netanel In the first on we have $$2+ frac{3}{x}-frac{35}{x^2}=2- frac{3}{5}-frac{35}{25}=frac{50-15-35}{25}=0$$
            – gimusi
            Nov 22 at 17:20




            1




            1




            For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
            – gimusi
            Nov 22 at 17:40




            For the second one $$x^2+ frac{3}{2}x-frac{35}{2}=frac12(2x-7)(x+5)$$
            – gimusi
            Nov 22 at 17:40


















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