I am really stuck with this question:
Suppose n is an odd positive integer. Prove that there exists a positive integer m such that (2^m − 1)\n .
(Here, “divides” means that when 2^m − 1 is divided by n.)
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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