The meaning of $(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$












0












$begingroup$


I'm trying to understand the meaning of this:



$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$



the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$






contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 7:11










  • $begingroup$
    What is "$R$"? The real numbers?
    $endgroup$
    – Eevee Trainer
    Dec 12 '18 at 7:11










  • $begingroup$
    R Mean real numbers(couldn't find the char sorry)
    $endgroup$
    – shay
    Dec 12 '18 at 7:15










  • $begingroup$
    I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
    $endgroup$
    – shay
    Dec 12 '18 at 7:17
















0












$begingroup$


I'm trying to understand the meaning of this:



$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$



the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$






contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 7:11










  • $begingroup$
    What is "$R$"? The real numbers?
    $endgroup$
    – Eevee Trainer
    Dec 12 '18 at 7:11










  • $begingroup$
    R Mean real numbers(couldn't find the char sorry)
    $endgroup$
    – shay
    Dec 12 '18 at 7:15










  • $begingroup$
    I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
    $endgroup$
    – shay
    Dec 12 '18 at 7:17














0












0








0





$begingroup$


I'm trying to understand the meaning of this:



$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$



the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$






contest-math







share|cite|improve this question











$endgroup$




I'm trying to understand the meaning of this:



$$(forall M in mathbb{R} )( exists B in mathbb{R} )( forall x>B )( f(x)<M )$$



the only thing I could figure out that if $xrightarrow infty$ then $f(x)$ not going to $+infty$






contest-math




contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 7:21









Eevee Trainer

5,7571936




5,7571936










asked Dec 12 '18 at 7:01









shayshay

103




103












  • $begingroup$
    Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 7:11










  • $begingroup$
    What is "$R$"? The real numbers?
    $endgroup$
    – Eevee Trainer
    Dec 12 '18 at 7:11










  • $begingroup$
    R Mean real numbers(couldn't find the char sorry)
    $endgroup$
    – shay
    Dec 12 '18 at 7:15










  • $begingroup$
    I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
    $endgroup$
    – shay
    Dec 12 '18 at 7:17


















  • $begingroup$
    Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 7:11










  • $begingroup$
    What is "$R$"? The real numbers?
    $endgroup$
    – Eevee Trainer
    Dec 12 '18 at 7:11










  • $begingroup$
    R Mean real numbers(couldn't find the char sorry)
    $endgroup$
    – shay
    Dec 12 '18 at 7:15










  • $begingroup$
    I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
    $endgroup$
    – shay
    Dec 12 '18 at 7:17
















$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11




$begingroup$
Try to set values of $M$ so that you can understand what it going on. For example, set $M = 0$. From the above, we get $B$ such that if $x > B$ then $f(x) < 0$. So $f(x)$ is eventually smaller than $0$ as $x to infty$. Now, set $M=-1$, then we get a $B$ such that if $x > B$ then $f(x) < -1$. So $f(x)$ is eventually smaller than $-1$ as $x to infty$. Now, change $-1$ to any arbitrary real number, and see if you can say something about $f$ from the given statement.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 7:11












$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11




$begingroup$
What is "$R$"? The real numbers?
$endgroup$
– Eevee Trainer
Dec 12 '18 at 7:11












$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15




$begingroup$
R Mean real numbers(couldn't find the char sorry)
$endgroup$
– shay
Dec 12 '18 at 7:15












$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17




$begingroup$
I still dont understand, bucause if I'll take x<B it could be anything, so i can't be bound
$endgroup$
– shay
Dec 12 '18 at 7:17










3 Answers
3






active

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0












$begingroup$

That's the definition of “$f(x) to -infty$ as $x to infty$”.



(At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.



    enter image description here






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.



      Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.



      That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

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        votes






        active

        oldest

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        0












        $begingroup$

        That's the definition of “$f(x) to -infty$ as $x to infty$”.



        (At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          That's the definition of “$f(x) to -infty$ as $x to infty$”.



          (At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            That's the definition of “$f(x) to -infty$ as $x to infty$”.



            (At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)






            share|cite|improve this answer









            $endgroup$



            That's the definition of “$f(x) to -infty$ as $x to infty$”.



            (At least for functions $f colon mathbf{R} to mathbf{R}$, i.e., functions defined for all $f in mathbf{R}$. If the domain is a subset of $mathbf{R}$, one has to modify the definition slightly.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 12 '18 at 7:49









            Hans LundmarkHans Lundmark

            35.5k564115




            35.5k564115























                1












                $begingroup$

                Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.



                enter image description here






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.



                  enter image description here






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$



                    Graphically, think of $M$ as any horizontal line in the standard coordinate plane. Then your sentence in question is saying that there exists some value $B$ (think of $B$ has a vertical line) such that for all values to the right of $B$, if you evaluate at $f$, they will be below the horizontal line you originally drew ($M$). As the picture below shows, you're essentially saying that 'past $B$, $f$ lives in some quadrant'.



                    enter image description here







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 12 '18 at 7:23









                    Eevee Trainer

                    5,7571936




                    5,7571936










                    answered Dec 12 '18 at 7:16









                    T. FoT. Fo

                    466311




                    466311























                        0












                        $begingroup$

                        This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.



                        Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.



                        That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.



                          Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.



                          That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.



                            Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.



                            That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.






                            share|cite|improve this answer









                            $endgroup$



                            This suggests that, after a certain point, $f$ is bounded above by some number, and to any $M$ there is a corresponding "starting point" for this behavior $B$.



                            Thus, you can choose any upper bound $M in mathbb{R}$. Then we have a corresponding bound $B$ on the argument of $f$. For all $x$ beyond that point, then, $f(x)$ is less than this upper bound.



                            That is, for any upper bound $M$ on $f$, there is a corresponding lower bound $B$ on $x$, such that for any $x > B$, $f(x) < M$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 7:19









                            Eevee TrainerEevee Trainer

                            5,7571936




                            5,7571936






























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