- \(\lim (ca_n) = ca\) for all \(c \in \mathbb{R}\);
- \(\lim (a_n + b_n) = a + b\);
- \(\lim (a_nb_n) = ab\);
- \(\lim (a_n/b_n) = a\);
Previous Definitions:
- Absolute value definition and other properties
- Sequence definitions and convergence definitions
- Subsequences
- Show the limit template
Problem Discussion
For (iii) we want to show that \(\lim (a_nb_n) = ab\). So we want to find \(N\) such that for any \(\epsilon > 0\),
The trick here to add and subtract a term \(ab_n\) and then use the triangle inequality:
Using the same trick from (ii), we can make \(\lvert a_n - a \rvert\) be as small as we want and similarly we can make \(\lvert b_n - b \rvert\) be as small as we want. The idea is to bound these two terms to values such that the total will be exactly \(\epsilon\) which is what we want to prove. So, given that \(\lim b_n = b\), then we know that for some \(n \geq N_1\),
Similarly, given that \(\lim a_n = a\), then we know that for some \(n \geq N_2\),
We still need to bound the term \(|b_n|\) since it’s not a constant. One trick is to use the convergent sequences theorem which says that all convergent sequences are bounded by some term \(M\). This means that we can replace the bound above with the following:
If we plug in everything back, we get
Finally we can choose \(N\) to be the maximum of \(N_1\) and \(N_2\) to guarantee the above bounds.
Formal Proof
Let \(\epsilon > 0\) be arbitrary. Since \(\lim a_n = a\) and \(\lim b_n = b\), both sequences are convergent and hence bounded by the convergent sequences theorem. Therefore, there exists a constant \(M\) such that \(|a_n| \leq M\) and \(|b_n| \leq M\) for all \(n\). Now, since \(\lim a_n = a\), there exists \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\),
where \(M\) is a constant such that \(|a_n| \leq M\) for all \(n\). Similarly, since \(\lim b_n = b\), there exists \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\),
Therefore, choose a natural number \(N\) satisfying \(\max(N_1,N_2)\). We now verify that this choice has the desired property. Let \(n \geq N\). Then,
as we wanted to show. \(\blacksquare\)