Let $F: [0, \infty) \to \mathbb{R}$ be the antiderivative of $f$ vanishing at $0$. Since $f$ takes on positive values, $F$ is strictly increasing, hence injective, and $\alpha = \sup \text{im } F > 0$, the supremum being considered on the extended line.

a) Fix a positive integer $n_0$ such that $n_0\alpha \ge 1$. If $n$ is an integer greater than $n_0$, then $F(x_n) = 1/n$ for some $x_n > 0$, by the intermediate value theorem. Uniqueness of $x_n$ follows from injectivity of $F$.

b) Since $F(x_{n+1}) = 1/(n+1) < 1/n = F(x_n)$, and $F$ is strictly increasing, $(x_n)_{n>n_0}$ is a strictly decreasing sequence of positive real numbers, so it is convergent. By continuity, $F(\lim_{n\to\infty} x_n) = \lim_{n\to\infty} F(x_n) = 0 = F(0)$, so $\lim_{n\to\infty} x_n = 0$, by injectivity of $F$. For each integer $n > n_0$, refer to the first mean value theorem to write $1/n = F(x_n) = x_n f(t_n)$, i.e., $nx_n = 1/f(t_n)$, for some positive $t_n < x_n$. Since $(x_n)_{n>n_0}$ converges to $0$, so does $(t_n)_{n>n_0}$. Consequently, $(nx_n)_{n>n_0}$ converges to $1/f(0)$, by continuity of $f$.