Sunday, 3 May 2015

calculus - textProve:,an>0,displaystylelimntoinftyan=LRightarrowdisplaystylelimntoinftyanfrac1n=1

I have this exercise to prove in calculus 1:



Prove:{an}>0,limn{an}=Llimn{an}1n=1



Where "{an}" is a sequence.




We proved in class that limn1n=0.



Any L0=1.



limn{an}=L,L0=1,limn1n=0limn{an}1n=L0=1QED



Is my solution wrong? The solution of the university is different.
Their solution is:




$\text{Because} \, a_n \to L \, \text{then for} \, \epsilon = \frac{L}{2} \, \text{exists} \, N \, \text{such that} \, \forall n>N: \\ \cfrac{L}{2} = L-\cfrac{L}{2}

No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...