limn→∞(13⋅8+⋯+16(2n−1)(3n+1))
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the squeeze theorem by putting the an between 1(2n−1)(2n+1) and 1(4n−1)(4n+1) but it doesn't work.
limn→∞(13⋅8+⋯+16(2n−1)(3n+1))
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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