I have this exercise to prove in calculus 1:
Prove:{an}>0,limn→∞{an}=L⇒limn→∞{an}1n=1
Where "{an}" is a sequence.
We proved in class that limn→∞1n=0.
Any L0=1.
limn→∞{an}=L,L0=1,limn→∞1n=0⇒limn→∞{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}
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