Pointwise limit function of a piecewise function












0












$begingroup$



Find the pointwise limit function of:
$$f_n(x)=begin{cases}
0 & |x|> 1/n \
nx+1 & x in [-1/n, 0) \
1-nx & x in [0, 1/n]
end{cases} $$




I think that in the limit, if we fix a certain $x$, we get that:



$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
infty & x in [0, 0) \
-infty & x in [0, 0]
end{cases} $$
Where the second line is an empty statement, we rewrite this to:
$$lim _{n to infty} f_n(x)=begin{cases}
0 & |x|> 0 \
-infty & x =0
end{cases} $$
Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$






real-analysis limits functions proof-verification pointwise-convergence







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    0












    $begingroup$



    Find the pointwise limit function of:
    $$f_n(x)=begin{cases}
    0 & |x|> 1/n \
    nx+1 & x in [-1/n, 0) \
    1-nx & x in [0, 1/n]
    end{cases} $$




    I think that in the limit, if we fix a certain $x$, we get that:



    $$lim _{n to infty} f_n(x)=begin{cases}
    0 & |x|> 0 \
    infty & x in [0, 0) \
    -infty & x in [0, 0]
    end{cases} $$
    Where the second line is an empty statement, we rewrite this to:
    $$lim _{n to infty} f_n(x)=begin{cases}
    0 & |x|> 0 \
    -infty & x =0
    end{cases} $$
    Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$






    real-analysis limits functions proof-verification pointwise-convergence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Find the pointwise limit function of:
      $$f_n(x)=begin{cases}
      0 & |x|> 1/n \
      nx+1 & x in [-1/n, 0) \
      1-nx & x in [0, 1/n]
      end{cases} $$




      I think that in the limit, if we fix a certain $x$, we get that:



      $$lim _{n to infty} f_n(x)=begin{cases}
      0 & |x|> 0 \
      infty & x in [0, 0) \
      -infty & x in [0, 0]
      end{cases} $$
      Where the second line is an empty statement, we rewrite this to:
      $$lim _{n to infty} f_n(x)=begin{cases}
      0 & |x|> 0 \
      -infty & x =0
      end{cases} $$
      Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$






      real-analysis limits functions proof-verification pointwise-convergence







      share|cite|improve this question









      $endgroup$





      Find the pointwise limit function of:
      $$f_n(x)=begin{cases}
      0 & |x|> 1/n \
      nx+1 & x in [-1/n, 0) \
      1-nx & x in [0, 1/n]
      end{cases} $$




      I think that in the limit, if we fix a certain $x$, we get that:



      $$lim _{n to infty} f_n(x)=begin{cases}
      0 & |x|> 0 \
      infty & x in [0, 0) \
      -infty & x in [0, 0]
      end{cases} $$
      Where the second line is an empty statement, we rewrite this to:
      $$lim _{n to infty} f_n(x)=begin{cases}
      0 & |x|> 0 \
      -infty & x =0
      end{cases} $$
      Did I do this correctly? I'm not sure what the formal argument should be to get rid of the half-open interval $[0,0)$






      real-analysis limits functions proof-verification pointwise-convergence




      real-analysis limits functions proof-verification pointwise-convergence






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      asked Jan 7 at 12:28









      Wesley StrikWesley Strik

      2,199424




      2,199424






















          3 Answers
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          Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.






          share|cite|improve this answer









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          • $begingroup$
            yes that makes sense
            $endgroup$
            – Wesley Strik
            Jan 7 at 12:52



















          1












          $begingroup$

          Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.



          You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
            But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.






            share|cite|improve this answer











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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                yes that makes sense
                $endgroup$
                – Wesley Strik
                Jan 7 at 12:52
















              2












              $begingroup$

              Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                yes that makes sense
                $endgroup$
                – Wesley Strik
                Jan 7 at 12:52














              2












              2








              2





              $begingroup$

              Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.






              share|cite|improve this answer









              $endgroup$



              Your answer is not correct. If $x=0$ then $f_n(x)=1$ for all $n$ and the limit is $1$. If $x neq 0$ then $f_n(x)=0$ for all $n$ sufficiently large so the limit is $0$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 7 at 12:32









              Kavi Rama MurthyKavi Rama Murthy

              74.6k53270




              74.6k53270












              • $begingroup$
                yes that makes sense
                $endgroup$
                – Wesley Strik
                Jan 7 at 12:52


















              • $begingroup$
                yes that makes sense
                $endgroup$
                – Wesley Strik
                Jan 7 at 12:52
















              $begingroup$
              yes that makes sense
              $endgroup$
              – Wesley Strik
              Jan 7 at 12:52




              $begingroup$
              yes that makes sense
              $endgroup$
              – Wesley Strik
              Jan 7 at 12:52











              1












              $begingroup$

              Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.



              You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.



                You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.



                  You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.






                  share|cite|improve this answer









                  $endgroup$



                  Consider $xneq 0$. Then there is an $n_0$ such that $|x|>1/n hspace{0.3cm}forall ngeq n_0 $: thus $f_n(x)=0$ definitively, and the pointwise limit is 0.



                  You are left only with $x=0$, but $f_n(0)=1 hspace{0.2cm}forall n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 12:35









                  F.BattistoniF.Battistoni

                  515




                  515























                      1












                      $begingroup$

                      Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
                      But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
                        But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
                          But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.






                          share|cite|improve this answer











                          $endgroup$



                          Your notations are odd. The conclusion that the limit be zero if $x$ is nonzero is correct. (But why would you write $|x|>0$ instead of $xneq 0$?)
                          But at zero, the limit is $1$. Just check that all functions have value $1$ at zero.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 7 at 12:59

























                          answered Jan 7 at 12:33









                          A. PongráczA. Pongrácz

                          6,0821929




                          6,0821929






























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