- #36
Bunny-chan
- 105
- 4
Let me see. I guess if we set [itex]m > 0, n = 1[/itex], we could say:[tex]C = \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \dots, \frac{9999}{10000}, \dots = 0.5, 0.6666..., 0.75, \dots, 0.9999, \dots[/tex]which constantly approximates from [itex]0,9999...[/itex]haruspex said:Yes, convergence is all to do with limits. Is that a problem? For the purposes of the question, I don't think you actually need to prove the convergence. Just show a sequence which pretty obviously converges to the extremum in your answer.
Or I think we can also say[tex]\lim\limits_{m \rightarrow \infty} \frac{m}{m + 1}=\lim\limits_{m \rightarrow \infty} \frac{1}{1+\frac{1}{m}}=\frac{1}{1 + \lim\limits_{m \rightarrow \infty} \frac{1}{m}}=1[/tex]Is that acceptable?