Prove $e^x \geq x+1$ I've tried finding a solution to this problem. I want to use the following definition of e
$e = \lim_{n \to\infty} (1+1/n)^n$
I have seen people argue with Bernoulli's Inequality, saying
$1+x \leq (1+x/n)^n \to e^x$
but how do I know $(1+x/n)^n \to e^x$? The arguments I've seen use continuity of the functions $x \to a^x$, where a is a real number, but the proofs I've seen of this use $e^x$ continuous and the proofs for $e^x$ continuous require $e^x >= x+1$... 
I'm thankful for any help ! :)
 A: Proof of the inequality using the convexity
Let $f(x)=e^x$ then we have $f''(x)=e^x\ge0$ so $f$ is a convex function and the equation of its tangent line at the point $x=0$ is $y=x+1$. What's the position of this tangent relative to the curve of $f$?
A: If $e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$, then
$$ e^x=\lim_{n\to\infty} \left(1+\frac1n\right)^{xn} $$
Let $n=\frac ux$ (which we can do so long as $x$ is constant):
$$\begin{align} e^x&=\lim_{u\to\infty} \left(1+\frac xu\right)^{u}\\
&=\lim_{u\to\infty}\sum_{k=0}^u{u\choose k}\left(\frac xu\right)^k\\
&=\lim_{u\to\infty}\sum_{k=0}^u{u\choose k}u^{-k}x^k\\
&=\lim_{u\to\infty}\sum_{k=0}^u \frac1{k!}\frac uu\frac {u-1}u \cdots\frac{u-k+1}ux^k\\
&=\sum_{k=0}^\infty\frac{x^k}{k!}\\
&=1+x+\sum_{k=2}^\infty\frac{x^n}{k!}>1+x\end{align}$$
A: At $x=0$ we have $e^x=x+1$ and the derivative of $e^x$ is $e^x$ while the derivative of $x+1$ is just 1. 
For $x>0$ we have $e^x>1$ since $\log(e^x)=x>\log(1)=0$. So $e^x$ grows faster than $x+1$ as $x$ increases. 
For $x<0$ we have $e^x<1$ for the same reasons so $e^x$ goes down slower than $x+1$ as $x$ decreases.
A: HINT:
$$\left(1 + \frac{1}{\frac {n}{x}} \right)^{\frac{n}{x}} = e\text{, as} \frac{n}{x} \to \infty$$
Raising both sides to $x$ will give you what you want. Then just apply Bernoulli's Inequality as you've mentioned.
