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

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

• Addressing the title, for all $x, e^{-x}\ge1-x$ so $x+e^{-x}\ge1>0$ – J. W. Tanner Mar 21 at 4:45
• 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. – Minus One-Twelfth Mar 21 at 4:46
• 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". – David Mar 21 at 4:49
• Thank you. That's what I was missing. Going backward. :( – Paichu Mar 21 at 4:50

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.

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.

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 $$x$$.

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: $$\color{blue}{e^x} = 1+x + \underbrace{\frac{e^{\xi}}{2}x^2}_{\geq 0} \color{blue}{> x}$$