how to prove limit exits using sequential limit definition

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Let $f:(-infty, 1) cup (1,infty) to mathbbR$ be defined by $f(x)=fracx^3-1x-1$ . Prove that $lim _xto 1f(x)=3$



i am proving this problem by limit of the definition



$|f(x)-3|=left|fracx^3-1x-1-3right|=|x^2+x-2|=|x-1||x+2|leq |x-1|(|x|+2)=4|x-1|<epsilon $ as i choose $|x-1|<delta=epsilon/4$



how to prove this by sequence of limit



ie how to prove by if $x_n to 1$ then $f(x_n)to 3$






sequences-and-series limits







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  • @farruhota how its dupicate i asked how to prove by sequence of limit
    – Inverse Problem
    Sep 4 at 12:19














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Let $f:(-infty, 1) cup (1,infty) to mathbbR$ be defined by $f(x)=fracx^3-1x-1$ . Prove that $lim _xto 1f(x)=3$



i am proving this problem by limit of the definition



$|f(x)-3|=left|fracx^3-1x-1-3right|=|x^2+x-2|=|x-1||x+2|leq |x-1|(|x|+2)=4|x-1|<epsilon $ as i choose $|x-1|<delta=epsilon/4$



how to prove this by sequence of limit



ie how to prove by if $x_n to 1$ then $f(x_n)to 3$






sequences-and-series limits







share|cite|improve this question





















  • @farruhota how its dupicate i asked how to prove by sequence of limit
    – Inverse Problem
    Sep 4 at 12:19












up vote
0
down vote

favorite









up vote
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down vote

favorite











Let $f:(-infty, 1) cup (1,infty) to mathbbR$ be defined by $f(x)=fracx^3-1x-1$ . Prove that $lim _xto 1f(x)=3$



i am proving this problem by limit of the definition



$|f(x)-3|=left|fracx^3-1x-1-3right|=|x^2+x-2|=|x-1||x+2|leq |x-1|(|x|+2)=4|x-1|<epsilon $ as i choose $|x-1|<delta=epsilon/4$



how to prove this by sequence of limit



ie how to prove by if $x_n to 1$ then $f(x_n)to 3$






sequences-and-series limits







share|cite|improve this question













Let $f:(-infty, 1) cup (1,infty) to mathbbR$ be defined by $f(x)=fracx^3-1x-1$ . Prove that $lim _xto 1f(x)=3$



i am proving this problem by limit of the definition



$|f(x)-3|=left|fracx^3-1x-1-3right|=|x^2+x-2|=|x-1||x+2|leq |x-1|(|x|+2)=4|x-1|<epsilon $ as i choose $|x-1|<delta=epsilon/4$



how to prove this by sequence of limit



ie how to prove by if $x_n to 1$ then $f(x_n)to 3$






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 4 at 11:39









Inverse Problem

906918




906918











  • @farruhota how its dupicate i asked how to prove by sequence of limit
    – Inverse Problem
    Sep 4 at 12:19
















  • @farruhota how its dupicate i asked how to prove by sequence of limit
    – Inverse Problem
    Sep 4 at 12:19















@farruhota how its dupicate i asked how to prove by sequence of limit
– Inverse Problem
Sep 4 at 12:19




@farruhota how its dupicate i asked how to prove by sequence of limit
– Inverse Problem
Sep 4 at 12:19










2 Answers
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Let it be given that $x_n to 1$. This means you can make $|x_n-1| < epsilon$, taking $n$ big enough.



We want to show $f(x_n) to 3$, so we want to show that we can make |$f(x_n)-3|<epsilon$, by taking large enough $n$.



But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.



All that's left is writing up all the details.






share|cite|improve this answer



























    up vote
    1
    down vote













    You already know that $f(x)-3=x^2+x-2$. If $x_n to 1$ then $x_n^2 to 1$ so $f(x)-3 to 1+1-2=0$ and $f(x) to 3$.






    share|cite|improve this answer




















    • ..its okay what if i want to prove both left and right limits exists and equal
      – Inverse Problem
      Sep 4 at 11:53










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Let it be given that $x_n to 1$. This means you can make $|x_n-1| < epsilon$, taking $n$ big enough.



    We want to show $f(x_n) to 3$, so we want to show that we can make |$f(x_n)-3|<epsilon$, by taking large enough $n$.



    But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.



    All that's left is writing up all the details.






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      Let it be given that $x_n to 1$. This means you can make $|x_n-1| < epsilon$, taking $n$ big enough.



      We want to show $f(x_n) to 3$, so we want to show that we can make |$f(x_n)-3|<epsilon$, by taking large enough $n$.



      But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.



      All that's left is writing up all the details.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Let it be given that $x_n to 1$. This means you can make $|x_n-1| < epsilon$, taking $n$ big enough.



        We want to show $f(x_n) to 3$, so we want to show that we can make |$f(x_n)-3|<epsilon$, by taking large enough $n$.



        But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.



        All that's left is writing up all the details.






        share|cite|improve this answer












        Let it be given that $x_n to 1$. This means you can make $|x_n-1| < epsilon$, taking $n$ big enough.



        We want to show $f(x_n) to 3$, so we want to show that we can make |$f(x_n)-3|<epsilon$, by taking large enough $n$.



        But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.



        All that's left is writing up all the details.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 4 at 12:00









        JuliusL33t

        1,260817




        1,260817




















            up vote
            1
            down vote













            You already know that $f(x)-3=x^2+x-2$. If $x_n to 1$ then $x_n^2 to 1$ so $f(x)-3 to 1+1-2=0$ and $f(x) to 3$.






            share|cite|improve this answer




















            • ..its okay what if i want to prove both left and right limits exists and equal
              – Inverse Problem
              Sep 4 at 11:53














            up vote
            1
            down vote













            You already know that $f(x)-3=x^2+x-2$. If $x_n to 1$ then $x_n^2 to 1$ so $f(x)-3 to 1+1-2=0$ and $f(x) to 3$.






            share|cite|improve this answer




















            • ..its okay what if i want to prove both left and right limits exists and equal
              – Inverse Problem
              Sep 4 at 11:53












            up vote
            1
            down vote










            up vote
            1
            down vote









            You already know that $f(x)-3=x^2+x-2$. If $x_n to 1$ then $x_n^2 to 1$ so $f(x)-3 to 1+1-2=0$ and $f(x) to 3$.






            share|cite|improve this answer












            You already know that $f(x)-3=x^2+x-2$. If $x_n to 1$ then $x_n^2 to 1$ so $f(x)-3 to 1+1-2=0$ and $f(x) to 3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 4 at 11:44









            Kavi Rama Murthy

            26.1k31437




            26.1k31437











            • ..its okay what if i want to prove both left and right limits exists and equal
              – Inverse Problem
              Sep 4 at 11:53
















            • ..its okay what if i want to prove both left and right limits exists and equal
              – Inverse Problem
              Sep 4 at 11:53















            ..its okay what if i want to prove both left and right limits exists and equal
            – Inverse Problem
            Sep 4 at 11:53




            ..its okay what if i want to prove both left and right limits exists and equal
            – Inverse Problem
            Sep 4 at 11:53

















             

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