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 $x > - 1$:
\(\ds e^x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \left({1 + \frac x n}\right)^n\) | Definition of Real Exponential Function | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \lim_{n \mathop \to \infty} \left({1 + x}\right)\) | Bernoulli's Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + x\) |
as required.
For $x \le -1$, the inequality follows from the fact that $e^x$ is positive for all $x$.
$\blacksquare$