Limit Query on $-infty$ or $infty$ answer











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Let $f(x)=x+log_e(x)-xlog_e(x)$
I am confused if I want to know whether $lim_{xrightarrow infty }f(x)$ is $-infty$ or $infty$



I used the following concept $f(x)=x+log_e(x)-xlog_e(x)$



$lim_{xrightarrow infty }f(x)$=$infty+log_e(infty)(1-infty)$.



After this step I am confused










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  • $f'<0$ then the function is decreasing for $x>1$.
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Let $f(x)=x+log_e(x)-xlog_e(x)$
I am confused if I want to know whether $lim_{xrightarrow infty }f(x)$ is $-infty$ or $infty$



I used the following concept $f(x)=x+log_e(x)-xlog_e(x)$



$lim_{xrightarrow infty }f(x)$=$infty+log_e(infty)(1-infty)$.



After this step I am confused










share|cite|improve this question






















  • $f'<0$ then the function is decreasing for $x>1$.
    – Nosrati
    yesterday















up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
2






2





Let $f(x)=x+log_e(x)-xlog_e(x)$
I am confused if I want to know whether $lim_{xrightarrow infty }f(x)$ is $-infty$ or $infty$



I used the following concept $f(x)=x+log_e(x)-xlog_e(x)$



$lim_{xrightarrow infty }f(x)$=$infty+log_e(infty)(1-infty)$.



After this step I am confused










share|cite|improve this question













Let $f(x)=x+log_e(x)-xlog_e(x)$
I am confused if I want to know whether $lim_{xrightarrow infty }f(x)$ is $-infty$ or $infty$



I used the following concept $f(x)=x+log_e(x)-xlog_e(x)$



$lim_{xrightarrow infty }f(x)$=$infty+log_e(infty)(1-infty)$.



After this step I am confused







limits






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asked yesterday









Samar Imam Zaidi

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  • $f'<0$ then the function is decreasing for $x>1$.
    – Nosrati
    yesterday




















  • $f'<0$ then the function is decreasing for $x>1$.
    – Nosrati
    yesterday


















$f'<0$ then the function is decreasing for $x>1$.
– Nosrati
yesterday






$f'<0$ then the function is decreasing for $x>1$.
– Nosrati
yesterday












3 Answers
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If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.



However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
$f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$



Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$



So, $lim_{xrightarrow infty }f(x) = - infty$






share|cite|improve this answer






























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    0
    down vote













    At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$



    We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.



    If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.



    So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$



    Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$



    Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$






    share|cite|improve this answer






























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      $$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.





      Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.






      share|cite|improve this answer























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        3 Answers
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        3 Answers
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        up vote
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        If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.



        However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
        $f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$



        Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$



        So, $lim_{xrightarrow infty }f(x) = - infty$






        share|cite|improve this answer



























          up vote
          0
          down vote













          If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.



          However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
          $f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$



          Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$



          So, $lim_{xrightarrow infty }f(x) = - infty$






          share|cite|improve this answer

























            up vote
            0
            down vote










            up vote
            0
            down vote









            If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.



            However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
            $f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$



            Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$



            So, $lim_{xrightarrow infty }f(x) = - infty$






            share|cite|improve this answer














            If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.



            However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
            $f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$



            Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$



            So, $lim_{xrightarrow infty }f(x) = - infty$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            Tom Collinge

            4,4641133




            4,4641133






















                up vote
                0
                down vote













                At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$



                We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.



                If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.



                So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$



                Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$



                Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$






                share|cite|improve this answer



























                  up vote
                  0
                  down vote













                  At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$



                  We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.



                  If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.



                  So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$



                  Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$



                  Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$






                  share|cite|improve this answer

























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$



                    We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.



                    If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.



                    So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$



                    Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$



                    Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$






                    share|cite|improve this answer














                    At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$



                    We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.



                    If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.



                    So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$



                    Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$



                    Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited yesterday

























                    answered yesterday









                    DanielWainfleet

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                        $$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.





                        Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.






                        share|cite|improve this answer



























                          up vote
                          0
                          down vote













                          $$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.





                          Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.





                            Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.






                            share|cite|improve this answer














                            $$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.





                            Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited yesterday

























                            answered yesterday









                            heropup

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