Question

Mathematical Real Analysis

Convergence Sequence Conception Question:

By the definition: for all epsilon >0 there exists a N such that for all n>N absolute an-a < epsilon

Question: assume bn ----->b and b is not 0. prove that lim n to infinity 1/ bn = 1/ b .

**Please Tell me why here we need to have N1 and N2 and
Find the Max(N1,N2) but other example we don't.**

**Solve the question step by step as well**

Answer #1

Using the definition of convergence of a sequence, prove that the
sequence converges to the proposed limit.
lim (as n goes to infinity) 1/(n^2) = 0

Question about the Mathematical Real Analysis Proof
Show that if xn → 0 then √xn → 0.
Proof. Let ε > 0 be arbitrary. Since xn → 0 there is some N
∈N such that |xn| < ε^2 for all n > N. Then for all n > N
we have that |√xn| < ε
My question is based on the sequence convergence
definition it should be absolute an-a<ε but here
why we can take xn<ε^2 rather than ε?...

Prove that if for epsilon >0 there exists a positive integer n
such that for all n>N we have p_n is an element of
(x+(-epsilon),x+epsilon) then p_1,p_2, ... p_n converges to
x.

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