Why is $x+e^{-x}>0$ for all $x in mathbb{R}$? The 2019 Stack Overflow Developer Survey...

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Why is $x+e^{-x}>0$ for all $x in mathbb{R}$?



The 2019 Stack Overflow Developer Survey Results Are InSimplest or nicest proof that $1+x le e^x$What is the limit of this function as $n$ tends to infinity? (something to the power of e)Limit question: $lim_{xto infty}f(x),;; lim_{xto -infty}f(x)$?Does $displaystylelim_{ntoinfty}frac{n}{n+1} = 1$?Calculus - Finding the rate of change of an edge of a cube.Calculus -as X approaches O or infinityProve limit exists for an almost monotonic bounded function, involves convexity and square integrabilityCan I not use L'hopital rule here?What do second order partial derivatives mean in graphs?Find the simplest counterexample against exchanging limit and summationIf $f : [a, infty) → Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_{x→infty} f(x) = 0$.












2












$begingroup$


Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?



I think I am missing something really simple here. Please help.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
    $endgroup$
    – J. W. Tanner
    Mar 21 at 4:45








  • 1




    $begingroup$
    To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 4:46








  • 4




    $begingroup$
    Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
    $endgroup$
    – David
    Mar 21 at 4:49








  • 1




    $begingroup$
    Thank you. That's what I was missing. Going backward. :(
    $endgroup$
    – Paichu
    Mar 21 at 4:50
















2












$begingroup$


Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?



I think I am missing something really simple here. Please help.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
    $endgroup$
    – J. W. Tanner
    Mar 21 at 4:45








  • 1




    $begingroup$
    To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 4:46








  • 4




    $begingroup$
    Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
    $endgroup$
    – David
    Mar 21 at 4:49








  • 1




    $begingroup$
    Thank you. That's what I was missing. Going backward. :(
    $endgroup$
    – Paichu
    Mar 21 at 4:50














2












2








2


2



$begingroup$


Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?



I think I am missing something really simple here. Please help.






calculus limits







share|cite|improve this question











$endgroup$




Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?



I think I am missing something really simple here. Please help.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 6:31









David G. Stork

12.2k41836




12.2k41836










asked Mar 21 at 4:42









PaichuPaichu

804616




804616












  • $begingroup$
    Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
    $endgroup$
    – J. W. Tanner
    Mar 21 at 4:45








  • 1




    $begingroup$
    To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 4:46








  • 4




    $begingroup$
    Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
    $endgroup$
    – David
    Mar 21 at 4:49








  • 1




    $begingroup$
    Thank you. That's what I was missing. Going backward. :(
    $endgroup$
    – Paichu
    Mar 21 at 4:50


















  • $begingroup$
    Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
    $endgroup$
    – J. W. Tanner
    Mar 21 at 4:45








  • 1




    $begingroup$
    To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 4:46








  • 4




    $begingroup$
    Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
    $endgroup$
    – David
    Mar 21 at 4:49








  • 1




    $begingroup$
    Thank you. That's what I was missing. Going backward. :(
    $endgroup$
    – Paichu
    Mar 21 at 4:50
















$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45






$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45






1




1




$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46






$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46






4




4




$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49






$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49






1




1




$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50




$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50










4 Answers
4






active

oldest

votes


















2












$begingroup$

You have got your directions mixed up.



When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Take



    $x < y < 0; tag 1$



    then



    $f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$



    it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that



    $s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$



    if we now choose



    $y < y_0, tag 4$



    then



    $f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$



    we re-arrange this inequality:



    $f(x) - f(y) > -M(y - x) = M(x - y), tag 6$



    $f(x) > f(y) + M(x - y); tag 7$



    we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,



    $displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$



    and hence



    $displaystyle lim_{x to -infty}f(x) = infty tag 9$



    as well.



    We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.



    Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show



    $x + e^{-x} > 0, forall x in Bbb R; tag{10}$



    for



    $x ge 0, tag{11}$



    $e^{-x} > 0 tag{12}$



    as well, hence we also have



    $x + e^{-x} > 0; tag{13}$



    for



    $x < 0, tag{14}$



    we may use the power series for $e^{-x}$:



    $e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$



    then



    $x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$



    since every term on the right is positive when $x < 0$. End of Note.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.



      Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Note that




        • $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$


        The last inequality follows directly by Taylor:
        $$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$






        share|cite|improve this answer









        $endgroup$














          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You have got your directions mixed up.



          When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            You have got your directions mixed up.



            When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              You have got your directions mixed up.



              When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.






              share|cite|improve this answer









              $endgroup$



              You have got your directions mixed up.



              When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 21 at 4:51









              астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

              40.3k33678




              40.3k33678























                  3












                  $begingroup$

                  Take



                  $x < y < 0; tag 1$



                  then



                  $f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$



                  it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that



                  $s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$



                  if we now choose



                  $y < y_0, tag 4$



                  then



                  $f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$



                  we re-arrange this inequality:



                  $f(x) - f(y) > -M(y - x) = M(x - y), tag 6$



                  $f(x) > f(y) + M(x - y); tag 7$



                  we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,



                  $displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$



                  and hence



                  $displaystyle lim_{x to -infty}f(x) = infty tag 9$



                  as well.



                  We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.



                  Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show



                  $x + e^{-x} > 0, forall x in Bbb R; tag{10}$



                  for



                  $x ge 0, tag{11}$



                  $e^{-x} > 0 tag{12}$



                  as well, hence we also have



                  $x + e^{-x} > 0; tag{13}$



                  for



                  $x < 0, tag{14}$



                  we may use the power series for $e^{-x}$:



                  $e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$



                  then



                  $x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$



                  since every term on the right is positive when $x < 0$. End of Note.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    Take



                    $x < y < 0; tag 1$



                    then



                    $f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$



                    it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that



                    $s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$



                    if we now choose



                    $y < y_0, tag 4$



                    then



                    $f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$



                    we re-arrange this inequality:



                    $f(x) - f(y) > -M(y - x) = M(x - y), tag 6$



                    $f(x) > f(y) + M(x - y); tag 7$



                    we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,



                    $displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$



                    and hence



                    $displaystyle lim_{x to -infty}f(x) = infty tag 9$



                    as well.



                    We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.



                    Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show



                    $x + e^{-x} > 0, forall x in Bbb R; tag{10}$



                    for



                    $x ge 0, tag{11}$



                    $e^{-x} > 0 tag{12}$



                    as well, hence we also have



                    $x + e^{-x} > 0; tag{13}$



                    for



                    $x < 0, tag{14}$



                    we may use the power series for $e^{-x}$:



                    $e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$



                    then



                    $x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$



                    since every term on the right is positive when $x < 0$. End of Note.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Take



                      $x < y < 0; tag 1$



                      then



                      $f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$



                      it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that



                      $s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$



                      if we now choose



                      $y < y_0, tag 4$



                      then



                      $f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$



                      we re-arrange this inequality:



                      $f(x) - f(y) > -M(y - x) = M(x - y), tag 6$



                      $f(x) > f(y) + M(x - y); tag 7$



                      we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,



                      $displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$



                      and hence



                      $displaystyle lim_{x to -infty}f(x) = infty tag 9$



                      as well.



                      We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.



                      Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show



                      $x + e^{-x} > 0, forall x in Bbb R; tag{10}$



                      for



                      $x ge 0, tag{11}$



                      $e^{-x} > 0 tag{12}$



                      as well, hence we also have



                      $x + e^{-x} > 0; tag{13}$



                      for



                      $x < 0, tag{14}$



                      we may use the power series for $e^{-x}$:



                      $e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$



                      then



                      $x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$



                      since every term on the right is positive when $x < 0$. End of Note.






                      share|cite|improve this answer











                      $endgroup$



                      Take



                      $x < y < 0; tag 1$



                      then



                      $f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$



                      it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that



                      $s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$



                      if we now choose



                      $y < y_0, tag 4$



                      then



                      $f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$



                      we re-arrange this inequality:



                      $f(x) - f(y) > -M(y - x) = M(x - y), tag 6$



                      $f(x) > f(y) + M(x - y); tag 7$



                      we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,



                      $displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$



                      and hence



                      $displaystyle lim_{x to -infty}f(x) = infty tag 9$



                      as well.



                      We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.



                      Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show



                      $x + e^{-x} > 0, forall x in Bbb R; tag{10}$



                      for



                      $x ge 0, tag{11}$



                      $e^{-x} > 0 tag{12}$



                      as well, hence we also have



                      $x + e^{-x} > 0; tag{13}$



                      for



                      $x < 0, tag{14}$



                      we may use the power series for $e^{-x}$:



                      $e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$



                      then



                      $x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$



                      since every term on the right is positive when $x < 0$. End of Note.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 21 at 5:46

























                      answered Mar 21 at 5:29









                      Robert LewisRobert Lewis

                      48.9k23168




                      48.9k23168























                          2












                          $begingroup$

                          Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.



                          Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.



                            Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.



                              Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.






                              share|cite|improve this answer









                              $endgroup$



                              Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.



                              Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Mar 21 at 4:51









                              copper.hatcopper.hat

                              128k561161




                              128k561161























                                  0












                                  $begingroup$

                                  Note that




                                  • $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$


                                  The last inequality follows directly by Taylor:
                                  $$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Note that




                                    • $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$


                                    The last inequality follows directly by Taylor:
                                    $$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Note that




                                      • $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$


                                      The last inequality follows directly by Taylor:
                                      $$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Note that




                                      • $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$


                                      The last inequality follows directly by Taylor:
                                      $$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Mar 21 at 7:03









                                      trancelocationtrancelocation

                                      13.7k1829




                                      13.7k1829






























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