Exponential of x not less than 1+x

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Theorem

$e^x \ge 1 + x$

for all $x \in \R$.


Proof

For $x > - 1$:

\(\displaystyle e^x\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac x n}\right)^n\) Definition of Real Exponential Function
\(\displaystyle \) \(\ge\) \(\displaystyle \lim_{n \mathop \to \infty} \left({1 + x}\right)\) Bernoulli's Inequality
\(\displaystyle \) \(=\) \(\displaystyle 1 + x\)

as required.

For $x \le -1$, the inequality follows from the fact that $e^x$ is positive for all $x$.

$\blacksquare$