# Proof

(related to Proposition: Product of Convegent Real Sequences)

Let the real sequences $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ be convergent with $$\lim_{n\rightarrow\infty} a_n=a$$ and $$\lim_{n\rightarrow\infty} b_n=b$$. It follows from the corresponding proposition that they are bounded by some positive real number $$B > 0$$. By making $$B$$ sufficiently large, and without any loss of generality, we can assume $$|a_n|\le B$$ and $$|b_n| \le B$$ for all $$n\in\mathbb N$$. Further, let $$\epsilon > 0$$. It follows that

1. there is an $$N(\epsilon,B)\in\mathbb N$$ with $$|a_n - a| < \frac{\epsilon}{2B}$$ for all $$n > N(\epsilon,B)$$, and that
2. there is an $$M(\epsilon,B)\in\mathbb N$$ with $$|b_n - b| < \frac{\epsilon}{2B}$$ for all $$n > M(\epsilon,B)$$.

Therefore, for all $$n > \max(N(\epsilon,B),M(\epsilon,B))$$ we get by virtue of the triangle inequality: $|a_nb_n - ab| = |a_n(b_n - b)+ (a_n-a)b| \le |a_n||b_n - b| + |a_n - a||b| < B\frac{\epsilon}{2B} + \frac{\epsilon}{2B}B=\epsilon.$

Since $$\epsilon$$ can be arbitrarily small chosen, it follows that

$\lim_{n\rightarrow\infty} (a_n \cdot b_n)=\lim_{n\rightarrow\infty} a_n \cdot \lim_{n\rightarrow\infty} b_n.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983