I have this exercise to prove in calculus 1:
$$ \text{Prove:} \, \{a_n\}>0, \displaystyle{\lim_{n \to \infty}} \{ a_n \}=L \Rightarrow \displaystyle{ \lim_{n \to \infty } \{a_n \}^\frac{1}{n}}=1 $$
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:
$\text{Because} \, a_n \to L \, \text{then for} \, \epsilon = \frac{L}{2} \, \text{exists} \, N \, \text{such that} \, \forall n>N: \\ \cfrac{L}{2} = L-\cfrac{L}{2}<a_n<L+\cfrac{L}{2}=\cfrac{3L}{2} \\ \text{and therefore:} \\ 1 \leftarrow \sqrt[n]{\cfrac{L}{2}}<\sqrt[n]{a_n}<\sqrt[n]{\cfrac{3L}{2}}\to 1$
Your reasoning is slightly wrong. To put it in the right way, state it as $$\lim_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty} e^{\frac{1}{n}\log(a_n)} = e^{\lim_{n \to \infty} \frac{1}{n}\log\left( \lim_{n \to \infty} a_n\right)} = e^{0\cdot L} = e^0 = 1$$ Since $\exp$ and $\log$ are continuous functions. However, the solution of the university uses much less additional knowledge and is therefore better. To point out the problem in your solution: You would have to proove first that $$\lim_{n \to \infty} a_n^{1/n} = \left(\lim_{n \to \infty} a_n \right)^{\lim_{n \to \infty} \frac{1}{n}}$$ holds.