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Theorem 1   If $0 \le x < 1$ , then$$ 1 + x \le \exp x \le {1 \over 1-x}$$

Proof. By the inequalities for differences of powers, we have$$ x \le \left( 1 + {x \over n} \right)^n - 1 \le {x \over 1 - \left( {n-1 \over n} \right) x} .$$ Since $n-1 < n$ , and $x > 0$ , we have $0 < (n-1 / n) x < x$ . Because $x < 1$ , this implies $1 - (n-1 / n) x > 1 - x$ , so$$ {x \over 1 - \left( {n-1 \over n} \right) x} < {x \over 1 - x}.$$ Hence$$ 1 + x \le \left( 1 + {x \over n} \right)^n \le {1 \over 1 - x}.$$ Taking the limit as $n \to \infty$ , we obtain our result.

Theorem 2   $$ \lim_{x \to 0} {\exp (x) - 1 \over x} = 1$$

Proof. Assume $0 < x < 1$ . By our bound, we have$$ 1 \le {\exp (x) - 1 \over x} \le {1 \over 1 - x} .$$ Suppose that $-1 < x < 0$ . Then, since $\exp (x) = 1 / \exp (-x)$ , we have$$ {\exp (x) - 1 \over x} = {1 \over \exp (-x)} \cdot {1 - \exp (-x) \over x} .$$ From the inequality above, we have$$ 1 \le {1 - \exp (-x) \over x} \le {1 \over 1+x} .$$ Hence$$ {1 \over \exp (-x)} \le {\exp (x) - 1 \over x} \le {1 \over (1+x) \exp (-x)}.$$ By theorem 1, we have $1 - x \le \exp (-x) \le 1 / (1 + x)$ , so$$ 1+x \le {1 - \exp (-x) \over x} \le {1 \over (1 + x)(1-x)} = {1 \over 1 - x^2} .$$ By the squeeze rule, we conclude that$$ \lim_{x \to 0} {1 - \exp (-x) \over x} = 1$$ whether we approach the limit from the left or the right.

Theorem 3   $$ {d \over dx} \exp (x) = \exp (x)$$

Proof. By definition,$$ {d \over dx} \exp (x) = \lim_{y \to x} {\exp (y) - \exp (x) \over y - x} .$$ By the addition theorem for the exponential, we have$$ {\exp (y) - \exp (x) \over y - x} = \exp (x) \cdot {\exp (y-x) - 1 \over y - x} ,$$ so$$ \lim_{y \to x} {\exp (y) - \exp (x) \over y - x} = \exp (x) \lim_{y \to x} {\exp (y-x) - 1 \over y - x} = \exp (x) \lim_{y \to 0} {\exp y - 1 \over y} .$$ By theorem 2, the limit on the right-hand side equals $1$ , so we have$$ \lim_{y \to x} {\exp (y) - \exp (x) \over y - x} = \exp (x) .$$

derivative of exponential function is owned by Raymond Puzio, Thomas Foregger, J. Pahikkala, Chi Woo.