Limit of a sequence involving absolute value and subsequences

Question

So I have the following question:

Suppose $lim|X_n|$ =3 but $X_n$ does not have a convergent sub-sequence with limit 3.

Prove $limX_n=-3$.

Here is my attempt:

1. $lim|X_n| =3$ so ${|X_n|}$ is bounded by the Boundedness Theorem. This implies ${X_n}$ is also bounded.

2. by properties of absolute value $-|a|\le a \le |a|$ so $-|X_n|\le X_n \le |X_n|$

3. taking the limit of inequality yields $-3 \le limX_n \le 3$

4. Suppose $limX_n$ = 3 contradiction because if $limX_n=x$ then all ${X_{n_k}}$ also converge to x and it is stated in the problem that there is no subsequence with limit 3.

5. Suppose $-3 \lt limX_n \lt 3$ which implies that $|limX_n|\lt3$ but $|limX_n|=lim|X_n|\lt3$ which is also contradiction because if $limX_n=x$ then $lim|X_n|=|X_n|$ and we know $lim|X_n|=3$

6. Thus $limX_n=3$

turned this in a few days ago and teacher was ok with it I just forgot to update post sorry

Answer 1

You can follow those directions :

• If $\lim\limits_{n \to \infty} \vert X_n \vert = 3$ then $(X_n)$ is bounded.
• As $(X_n)$ is bounded, it has at least a limit point.
• The only possible limit point is $3$.

Conclude.

Answer 2

Could we use the negation of the theorem: X = (xn ) be a bounded sequence of real numbers. Then we have the property that every convergent subsequence of X converges to x. Then the sequence X converges to x.
Then there exists a convergent subsequence of X does not convergent to x. Then the sequence X does not convergent to x. So lim(Xn) not equal to 3. When xn>0, lim(xn)=3, when xn<=0, lim(-xn)=-lim(xn)=3 then lim(xn)=-3. since lim(xn)not equal to 3, then xn not >0, then xn<=0, and lim(xn) =-3.