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Why is $x+e^-x>0$ for all $x in mathbbR$?

Simplest 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: $lim_xto inftyf(x),;; lim_xto -inftyf(x)$?Does $displaystylelim_ntoinftyfracnn+1 = 1$?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

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

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Addressing the title, for all $x, e^-xge1-x$ so $x+e^-xge1>0$
  $endgroup$
  – 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 $e^tge t+1$ for all real $t$. So $e^-xge -x+1> -x$ 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
  

  
  
  
  

add a comment | 

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

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Addressing the title, for all $x, e^-xge1-x$ so $x+e^-xge1>0$
  $endgroup$
  – 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 $e^tge t+1$ for all real $t$. So $e^-xge -x+1> -x$ 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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

$endgroup$

share|cite|improve this question

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

share|cite|improve this question

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

edited Mar 21 at 6:31

David G. Stork

11.5k41534

edited Mar 21 at 6:31

David G. Stork

11.5k41534

asked Mar 21 at 4:42

PaichuPaichu

790616

asked Mar 21 at 4:42

PaichuPaichu

790616

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

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

add a comment | 

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

$displaystyle lim_x to -inftyf(x) = infty tag 9$

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

$x + e^-x > 0, forall x in Bbb R; tag10$

for

$x ge 0, tag11$

$e^-x > 0 tag12$

as well, hence we also have

$x + e^-x > 0; tag13$

for

$x < 0, tag14$

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

$e^-x = 1 - x + dfracx^22! - dfracx^33 + ldots; tag15$

then

$x + e^-x = 1 + dfracx^22! - dfracx^33! + ldots > 0, tag16$

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

share|cite|improve this answer

edited Mar 21 at 5:46

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

$endgroup$

add a comment | 

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

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

0

$begingroup$

Note that

• $forall x in mathbbR : x+e^-x>0 Leftrightarrow forall x in mathbbR :e^-x>-x Leftrightarrow forall x in mathbbR :colorblueboxede^x>x$

The last inequality follows directly by Taylor:
$$colorbluee^x = 1+x + underbracefrace^xi2x^2_geq 0 colorblue> x$$

share|cite|improve this answer

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

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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

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

add a comment | 

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

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

$endgroup$

share|cite|improve this answer

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

answered Mar 21 at 4:51

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.2k33577

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

$displaystyle lim_x to -inftyf(x) = infty tag 9$

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

$x + e^-x > 0, forall x in Bbb R; tag10$

for

$x ge 0, tag11$

$e^-x > 0 tag12$

as well, hence we also have

$x + e^-x > 0; tag13$

for

$x < 0, tag14$

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

$e^-x = 1 - x + dfracx^22! - dfracx^33 + ldots; tag15$

then

$x + e^-x = 1 + dfracx^22! - dfracx^33! + ldots > 0, tag16$

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

share|cite|improve this answer

edited Mar 21 at 5:46

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

$endgroup$

add a comment | 

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

$displaystyle lim_x to -inftyf(x) = infty tag 9$

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

$x + e^-x > 0, forall x in Bbb R; tag10$

for

$x ge 0, tag11$

$e^-x > 0 tag12$

as well, hence we also have

$x + e^-x > 0; tag13$

for

$x < 0, tag14$

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

$e^-x = 1 - x + dfracx^22! - dfracx^33 + ldots; tag15$

then

$x + e^-x = 1 + dfracx^22! - dfracx^33! + ldots > 0, tag16$

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

share|cite|improve this answer

edited Mar 21 at 5:46

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

$endgroup$

add a comment | 

$begingroup$

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

$displaystyle lim_x to -inftyf(x) = infty tag 9$

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

$x + e^-x > 0, forall x in Bbb R; tag10$

for

$x ge 0, tag11$

$e^-x > 0 tag12$

as well, hence we also have

$x + e^-x > 0; tag13$

for

$x < 0, tag14$

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

$e^-x = 1 - x + dfracx^22! - dfracx^33 + ldots; tag15$

then

$x + e^-x = 1 + dfracx^22! - dfracx^33! + ldots > 0, tag16$

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

share|cite|improve this answer

edited Mar 21 at 5:46

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

$endgroup$

share|cite|improve this answer

edited Mar 21 at 5:46

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

answered Mar 21 at 5:29

Robert LewisRobert Lewis

48.6k23167

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

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

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

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

$endgroup$

share|cite|improve this answer

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

answered Mar 21 at 4:51

copper.hatcopper.hat

128k561161

0

$begingroup$

Note that

• $forall x in mathbbR : x+e^-x>0 Leftrightarrow forall x in mathbbR :e^-x>-x Leftrightarrow forall x in mathbbR :colorblueboxede^x>x$

The last inequality follows directly by Taylor:
$$colorbluee^x = 1+x + underbracefrace^xi2x^2_geq 0 colorblue> x$$

share|cite|improve this answer

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

0

$begingroup$

Note that

• $forall x in mathbbR : x+e^-x>0 Leftrightarrow forall x in mathbbR :e^-x>-x Leftrightarrow forall x in mathbbR :colorblueboxede^x>x$

The last inequality follows directly by Taylor:
$$colorbluee^x = 1+x + underbracefrace^xi2x^2_geq 0 colorblue> x$$

share|cite|improve this answer

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

$endgroup$

add a comment | 

$begingroup$

Note that

• $forall x in mathbbR : x+e^-x>0 Leftrightarrow forall x in mathbbR :e^-x>-x Leftrightarrow forall x in mathbbR :colorblueboxede^x>x$

The last inequality follows directly by Taylor:
$$colorbluee^x = 1+x + underbracefrace^xi2x^2_geq 0 colorblue> x$$

share|cite|improve this answer

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

$endgroup$

share|cite|improve this answer

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828

answered Mar 21 at 7:03

trancelocationtrancelocation

13.5k1828