I have this exercise to prove in calculus 1:
Prove:{an}>0,lim
Where "\{a_n\}" is a sequence.
We proved in class that \displaystyle{\lim_{{n \to \infty }} \cfrac{1}{n}} = 0 .
Any L^0=1.
\displaystyle{\lim_{n \to \infty}} \{ a_n \}=L,L^0=1, \displaystyle{\lim_{n \to \infty}} \frac{1}{n} = 0 \Rightarrow \\ \displaystyle{\lim_{n \to \infty}} \{ a_n \} ^\frac{1}{n} = L^0=1 \\ QED
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