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











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












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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009362%2fsolve-lim-x-rightarrow-5-frac2x2-502x23x-35%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009362%2fsolve-lim-x-rightarrow-5-frac2x2-502x23x-35%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more