Prove that a function satisfying given properties must be ex

Given a function $f(x)$ satisfying the properties:

$f(x+y) = f(x) f(y) \qquad \text{for all } x, y \in \mathbb{R}$

and

$f(x) = 1 + x g(x) \quad \text{where} \quad \lim_{x \to 0} g(x) = 1.$

Prove the following:

This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).

Proof. To show that the derivative $f'(x)$ exists for all $x$ we must show that the limit
$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

exists for all $x$ . Using the given properties of $f(x)$ we can evaluate this limit

$\begin{align*} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} &= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(x)(hg(h))}{h} &(\text{using } f(h) = 1 + hg(h)) \\[9pt] &= \lim_{h \to 0} f(x)g(h) \\[9pt] &= f(x) &(\text{using } \lim_{h \to 0} g(h) = 1). \end{align*}$

Therefore, $f'(x) = f(x)$ for all $x$ , so the derivative is defined everywhere $. \qquad \blacksquare$
Proof. From part (a) we know $f(x) = f'(x)$ . By Section 6.17, Exercise #39 (linked above) we know that the only functions $f(x)$ which satisfy this equation are $f(x) = 0$ for all $x$ or $f(x) = Ke^x$ for some constant $K$ (where $c = 1$ in the linked exercise). However, since the derivative of $f$ exists everywhere, and differentiability implies continuity, we know $f$ is continuous everywhere. Hence, $\lim_{x \to 0} f(x) = f(0)$ . Then,
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (1 + xg(x)) = 1 + 0 = 1$

since $\lim_{x \to 0} g(x) = 1$ , so $\lim_{x \to 0} xg(x) = 0$ . Therefore, we must have $f(x) = Ke^x$ for some constant $K$ . Furthermore, we must have $K = 1$ since $f(0) = Ke^0 = K = 1$ . Thus, $f(x) = e^x. \qquad \blacksquare$