Evaluate limn→∞sin2(π√(n!)2−(n!))
I tried it by Stirling's Approximation
n!≈√2πn.nne−n
but it leads us to nowhere.
Any hint will be of great help.
Answer
sin(π√(n!)2−n!)=sin(π√(n!)2−n!−π⋅n!)=sin(π−n!√(n!)2−n!+n!)=sin(π−1√1−1n!+1)→−sinπ2=−1
Evaluate limn→∞sin2(π√(n!)2−(n!))
I tried it by Stirling's Approximation
n!≈√2πn.nne−n
Any hint will be of great help.
Answer
sin(π√(n!)2−n!)=sin(π√(n!)2−n!−π⋅n!)=sin(π−n!√(n!)2−n!+n!)=sin(π−1√1−1n!+1)→−sinπ2=−1
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
No comments:
Post a Comment