Answer to Question #175367 in Real Analysis for Anand

Question

nLet (a

Let (a_n) n belongs to N be any sequence. Show that lim_n--->∞ a_n= L iff for every ε > 0 there exists some N belongs to N

such that n ≥ N implies a_n belongs to N_ε (L)

Answer 1

Solution:

Let $\left\{a_{n}\right\}$ be a sequence with positive terms such that $\lim _{n \rightarrow \infty} a_{n}=L>0$ .

 Let $x$ be a real number.

To prove that $\lim _{n \rightarrow \infty} a_{n}^{x}=L^{x}$ .

Proof: Let $\epsilon>0$. Note that $L<\left(L^{x}+\epsilon\right)^{1 / x}$ and $L>\left(L^{x}-\epsilon\right)^{1 / x}$ .

Since $\lim _{n \rightarrow \infty} a_{n}=L$ , there exists some N such that $n \geq N \implies a_{n}<\left(L^{x}+\epsilon\right)^{1 / x}$

and $a_{n}>\left(L^{x}-\epsilon\right)^{1 / x}$

Hence, for $n \geq N$ we have $a_{n}^{x}<L^{x}+\epsilon$

and $a_{n}^{x}>L^{x}-\epsilon$ . This shows $\lim _{n \rightarrow \infty} a_{n}^{x}=L^{x}$ .

Hence, proved.