Proving $\lim x_{n} = -10$. I am doing a problem which reads as follows.

Assume $\lim_{n \to \infty} |x_{n}| = 10$ but ${x_{n}}$ does not have a subsequence with limit $10$. Show that $\lim_{n \to \infty} x_{n} = -10$.

I know intuitively that the outcome must be true, but am having trouble proving this rigorously. I know that $x_{n}$ is bounded, and therefore it has a convergent subsequence by Bolzano-Weierstrass, would it be enough to say that because $|x_{n}|$ converges to $10$, $x_{n}$ must have a convergent subsequence to either $10$ or $-10$ by Bolzano, and because it doesn't converge to $10$ it must converge to $-10?$
 A: We use the result from this question which tells us that:

Let $(x_n)$ be a sequence of real numbers such that every subsequence of $(x_n)$ has a further subsequence which converges to $x$. Then the sequence $(x_n)$ converges to $x$.


We show that the hypothesis of the fact is true in our case.
Take an arbitrary subsequence $(x_{n_k})$ of the sequence. As you noted, this subsequence is bounded and hence, has a convergent (sub)subsequence $(x_{n_{k_m}})$. As you noted, the limit of this (sub)subsequence is either $10$ or $-10$. However, the former isn't possible. Thus, the (sub)sequence converges to $-10$.
Thus, $x = -10$ in the fact and we are done.
A: We don't need all the machinery in the other answer. As you mentioned, this feels intuitively clear and there is indeed a more intuitive and direct approach to it.

Claim. There are only finitely many $n$ such that $x_{n} > 0$.
Proof. Suppose not. Then, we have a subsequence $(x_{n_{k}})_{k}$ of $(x_{n})_{n}$ such that $x_{n_{k}} > 0$ for all $k$. In this case, we have $$\lim_{k \to \infty} x_{n_{k}} = \lim_{k \to \infty} |x_{n_{k}}| = 10,$$
a contradiction. $\qquad \square$
Thus, there exists $N > 0$ such that $x_{n} \leqslant 0$ for all $n \geqslant N$. Since ignoring finitely many terms does not affect convergence, we have $$\lim_{n \to \infty} x_{n} = \lim_{n \to \infty} -|x_{n}| = -10.$$
