Final answer:
To prove that the limit of the product of two sequences is the product of their limits, we find an N such that the sequences are within a small margin of their limits and use the triangle inequality to show that the absolute difference between the product of the sequences and the product of their limits is less than any given ε.
Step-by-step explanation:
To prove that lim n→∞ xnyn = LM, where xn → L and yn → M, we need to show that for any given ε > 0, there exists an N such that for all n ≥ N, |xnyn - LM| < ε. As xn approaches L and yn approaches M, let us consider ε > 0 and find an N such that |xn - L| < ε/(2M) and |yn - M| < ε/(2L) for all n ≥ N. This ensures that the sequences are within a small margin of their respective limits.
Now, we can express xnyn - LM as xnyn - L*yn + L*yn - LM, which can be rearranged to (xn - L)yn + L(yn - M). Applying the triangle inequality and substituting the established inequalities that bound xn and yn, we get:
|xnyn - LM| ≤ |xn - L||yn| + |L||yn - M| < (ε/(2M))*M + L*(ε/(2L)) = ε/2 + ε/2 = ε,
which establishes the product limit as LM. This proves that the product of the limits of the two sequences is equal to the limit of the product of the sequences as n approaches infinity.