Find limit of #lim 1^n/(2n-1)#?

Question

How do you find the limit of this? I got #0# as the answer but apparently that's not the answer.
#sum 1^n/(2n-1)#
#lim 1^n/(2n-1) = 1/oo = 0#

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Answer 1 · 2017-11-21T21:36:45

See below.

I am assuming this is a limit to infinity, if not then this is completely wrong.

#lim_(n->oo)(1^n/(2n-1))#

#((2n+1)*1^n)/((2n+1)(2n-1))=((2n+1)*1^n)/(4n^2-1)=(2n+1)/(4n^2-1)#

( for limits to infinity we can disregard the constants )

#->=(2n)/(4n^2)=1/(2n)#

as #x->oo# , #color(white)(88)1/(2n)->0#

#:.#

#lim_(n->oo)(1^n/(2n-1))=0#

So the limit is 0. Whoever told you it wasn't is wrong.