We have to show that $f_n(x)=x^ne^{-nx}$.











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So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?



Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.



Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.



My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?










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    up vote
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    down vote

    favorite












    So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?



    Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.



    Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.



    My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?



      Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.



      Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.



      My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?










      share|cite|improve this question













      So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?



      Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.



      Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.



      My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?







      real-analysis






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      asked Nov 14 at 13:20









      kim wenasa

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          Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.






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          • Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
            – kim wenasa
            Nov 14 at 14:10












          • That's another question. I suggest that you post it as such.
            – José Carlos Santos
            Nov 14 at 14:13











          Your Answer





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          Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.






          share|cite|improve this answer





















          • Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
            – kim wenasa
            Nov 14 at 14:10












          • That's another question. I suggest that you post it as such.
            – José Carlos Santos
            Nov 14 at 14:13















          up vote
          1
          down vote













          Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.






          share|cite|improve this answer





















          • Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
            – kim wenasa
            Nov 14 at 14:10












          • That's another question. I suggest that you post it as such.
            – José Carlos Santos
            Nov 14 at 14:13













          up vote
          1
          down vote










          up vote
          1
          down vote









          Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.






          share|cite|improve this answer












          Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 14 at 13:28









          José Carlos Santos

          140k18111203




          140k18111203












          • Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
            – kim wenasa
            Nov 14 at 14:10












          • That's another question. I suggest that you post it as such.
            – José Carlos Santos
            Nov 14 at 14:13


















          • Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
            – kim wenasa
            Nov 14 at 14:10












          • That's another question. I suggest that you post it as such.
            – José Carlos Santos
            Nov 14 at 14:13
















          Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
          – kim wenasa
          Nov 14 at 14:10






          Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
          – kim wenasa
          Nov 14 at 14:10














          That's another question. I suggest that you post it as such.
          – José Carlos Santos
          Nov 14 at 14:13




          That's another question. I suggest that you post it as such.
          – José Carlos Santos
          Nov 14 at 14:13


















           

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