$e^x>x\quad \forall x\in \Bbb R$ I'm trying to justify that $e^x> x\quad \forall x\in \Bbb R$:

It's clear but I want to give a formal argument. The only thing I've got is that: $$\text{Since}\lim_{x\to \infty} \frac{x}{e^x}=0\Rightarrow \exists M>0\; \text{such that}\; \frac{x}{e^x}\le M\; \Rightarrow x\le Me^x$$
and got stuck here.
Any ideas how to approach this?
 A: Since you have defined $e^x$ as the inverse of $\log x=\int_1^x\frac{dt}t$, it will be easiest to show that
$$ x>\log x$$
for all $x>0$. Then $e^x>x$ will follow by applying the exponential function on both sides (Hopefully you already know it is strictly increasing). And when $x\le 0$ we automatically have $e^x>x$ because $e^x$ is always positive.
Thus we're looking for
$$ x > \int_1^x \frac1t dt $$
When $x\ge 1$ this is easy, because then $1/t \le 1$ in the entire domain of integration, so $$\int_1^x \frac1t\, dt \le \int_1^x 1\,dt = x-1 < x $$
On the other hand, when $0<x<1$ the integral is negative and so clearly less than $x$!
A: Here's an outline in the form of hints: 


*

*Note that $e^x$ is non-negative, so $e^x > x$ for $x \in (-\infty,0)$.

*Examine the function $f(x) = e^x - x$.  What can we say about $f'(0)$?  What cay we say about $f'(x)$ for$x \in [0,\infty)$ in general?

*What does this imply about $f$ on $[0,\infty)$?

A: It really does depend on what you know. What if, for example, your definition of $e^x$? What are you allowed to use (like derivatives)?
If $e^x$ is defined as
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \dots, 
$$
it is almost clear that $e^x > x$ for all $x$.
If your definition is
$$
e^x = \lim_{n\to \infty}\left(1 + \frac{x}{n}\right)^n
$$
then, again, it is not too hard to see that $e^x > x$. 
A: The function $f(x) = e^x$ is convex.  The line $y = x + 1$ is the tangent line to this function's graph at $x = 0$.  Therefore,
$$e^x \ge x + 1 > x$$
for all real $x$.
