Limit of difference of sequence both going to $+infty$

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Let $(a_n), (b_n)$ be two sequences sucht that $a_n rightarrow +infty$ and $b_n rightarrow -infty$. Assume that for all $epsilon > 0$ we have $$fraca_nn leq epsilon$$ for $n$ large enough and for some $xi geq 0$$$fracb_n-n geq xi$$ for $n$ large enough.
How can I see that then the sequence $$C cdot a_n + D cdot b_n rightarrow -infty$$
for some constants $C,D > 0$?



Thanks a lot in advance!




calculus real-analysis analysis limits




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  • Sure, I wrote that a bit too fast.
    – vaoy
    Jul 30 at 7:45














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Let $(a_n), (b_n)$ be two sequences sucht that $a_n rightarrow +infty$ and $b_n rightarrow -infty$. Assume that for all $epsilon > 0$ we have $$fraca_nn leq epsilon$$ for $n$ large enough and for some $xi geq 0$$$fracb_n-n geq xi$$ for $n$ large enough.
How can I see that then the sequence $$C cdot a_n + D cdot b_n rightarrow -infty$$
for some constants $C,D > 0$?



Thanks a lot in advance!




calculus real-analysis analysis limits




share|cite|improve this question





















  • Sure, I wrote that a bit too fast.
    – vaoy
    Jul 30 at 7:45












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $(a_n), (b_n)$ be two sequences sucht that $a_n rightarrow +infty$ and $b_n rightarrow -infty$. Assume that for all $epsilon > 0$ we have $$fraca_nn leq epsilon$$ for $n$ large enough and for some $xi geq 0$$$fracb_n-n geq xi$$ for $n$ large enough.
How can I see that then the sequence $$C cdot a_n + D cdot b_n rightarrow -infty$$
for some constants $C,D > 0$?



Thanks a lot in advance!




calculus real-analysis analysis limits




share|cite|improve this question













Let $(a_n), (b_n)$ be two sequences sucht that $a_n rightarrow +infty$ and $b_n rightarrow -infty$. Assume that for all $epsilon > 0$ we have $$fraca_nn leq epsilon$$ for $n$ large enough and for some $xi geq 0$$$fracb_n-n geq xi$$ for $n$ large enough.
How can I see that then the sequence $$C cdot a_n + D cdot b_n rightarrow -infty$$
for some constants $C,D > 0$?



Thanks a lot in advance!





calculus real-analysis analysis limits





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share|cite|improve this question




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edited Jul 30 at 7:45
























asked Jul 30 at 7:03









vaoy

45228




45228











  • Sure, I wrote that a bit too fast.
    – vaoy
    Jul 30 at 7:45
















  • Sure, I wrote that a bit too fast.
    – vaoy
    Jul 30 at 7:45















Sure, I wrote that a bit too fast.
– vaoy
Jul 30 at 7:45




Sure, I wrote that a bit too fast.
– vaoy
Jul 30 at 7:45










2 Answers
2






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up vote
2
down vote



accepted










Let $xi >0$ and $epsilon = xi/2$, then, for $n$ large enough :



$$frac a_n + b_nn le xi/2-xi = -frac xi 2.$$



Then $a_n+b_n le -nfrac xi 2 to -infty$.






share|cite|improve this answer




























    up vote
    0
    down vote













    If there is $N in mathbb N$ such that



    $fraca_nn leq epsilon$ for all $ epsilon >0$ and all $n>N$, then we have



    $fraca_nn leq 0$ for all $n>N$, hence $a_n leq 0$ for all $n>N$.



    But then it is impossible to have $a_n rightarrow +infty$ !






    share|cite|improve this answer

















    • 1




      I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
      – nicomezi
      Jul 30 at 7:12










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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    up vote
    2
    down vote



    accepted










    Let $xi >0$ and $epsilon = xi/2$, then, for $n$ large enough :



    $$frac a_n + b_nn le xi/2-xi = -frac xi 2.$$



    Then $a_n+b_n le -nfrac xi 2 to -infty$.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Let $xi >0$ and $epsilon = xi/2$, then, for $n$ large enough :



      $$frac a_n + b_nn le xi/2-xi = -frac xi 2.$$



      Then $a_n+b_n le -nfrac xi 2 to -infty$.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Let $xi >0$ and $epsilon = xi/2$, then, for $n$ large enough :



        $$frac a_n + b_nn le xi/2-xi = -frac xi 2.$$



        Then $a_n+b_n le -nfrac xi 2 to -infty$.






        share|cite|improve this answer













        Let $xi >0$ and $epsilon = xi/2$, then, for $n$ large enough :



        $$frac a_n + b_nn le xi/2-xi = -frac xi 2.$$



        Then $a_n+b_n le -nfrac xi 2 to -infty$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 30 at 7:09









        nicomezi

        3,3871819




        3,3871819




















            up vote
            0
            down vote













            If there is $N in mathbb N$ such that



            $fraca_nn leq epsilon$ for all $ epsilon >0$ and all $n>N$, then we have



            $fraca_nn leq 0$ for all $n>N$, hence $a_n leq 0$ for all $n>N$.



            But then it is impossible to have $a_n rightarrow +infty$ !






            share|cite|improve this answer

















            • 1




              I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
              – nicomezi
              Jul 30 at 7:12














            up vote
            0
            down vote













            If there is $N in mathbb N$ such that



            $fraca_nn leq epsilon$ for all $ epsilon >0$ and all $n>N$, then we have



            $fraca_nn leq 0$ for all $n>N$, hence $a_n leq 0$ for all $n>N$.



            But then it is impossible to have $a_n rightarrow +infty$ !






            share|cite|improve this answer

















            • 1




              I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
              – nicomezi
              Jul 30 at 7:12












            up vote
            0
            down vote










            up vote
            0
            down vote









            If there is $N in mathbb N$ such that



            $fraca_nn leq epsilon$ for all $ epsilon >0$ and all $n>N$, then we have



            $fraca_nn leq 0$ for all $n>N$, hence $a_n leq 0$ for all $n>N$.



            But then it is impossible to have $a_n rightarrow +infty$ !






            share|cite|improve this answer













            If there is $N in mathbb N$ such that



            $fraca_nn leq epsilon$ for all $ epsilon >0$ and all $n>N$, then we have



            $fraca_nn leq 0$ for all $n>N$, hence $a_n leq 0$ for all $n>N$.



            But then it is impossible to have $a_n rightarrow +infty$ !







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 30 at 7:09









            Fred

            37k1237




            37k1237







            • 1




              I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
              – nicomezi
              Jul 30 at 7:12












            • 1




              I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
              – nicomezi
              Jul 30 at 7:12







            1




            1




            I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
            – nicomezi
            Jul 30 at 7:12




            I guess he wanted to say that for all $epsilon > 0$, there exists $N$ such that $forall nge N$, we have $frac a_n n le epsilon$.
            – nicomezi
            Jul 30 at 7:12












             

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