Prove if $x_n \leq k$ $\forall n \in N$ where $k$ is some constant integer, and $x_n \rightarrow x$, then $x \leq k$.

Question

So we know that $|x_n - x| < \varepsilon$ for all $\varepsilon > 0$ for some $n \geq N$. Since $x_n \leq k$, I think we can say that $|x_n - x| \leq |k - x| < \varepsilon$ for all $\varepsilon > 0$, but not entirely sure about this. I'm not sure what else I can conclude from here, but it looks like I can say that $|k - x| = |x - k| < \varepsilon$, so $k - \varepsilon < x < k + \varepsilon$. But we want $x \leq k$, and it says here that $x < k + \varepsilon$. How can I get rid of the $\varepsilon$ here?

Answer 1

Suppose, to the contrary, that $x > k$. Let $\epsilon := (x - k)/2$. Then $\epsilon > 0$, so since $x_n \to x$, there exists a positive integer $N$ such that $|x_N - x| < \epsilon$. In particular, $x_N > x - \epsilon = (k + x)/2 > k$, contradicting the assumption that $x_n \le k$ for all $n \in \Bbb N$.

Answer 2

Suppose that $x>k$, ie $x-k = \alpha$. For any $\epsilon>0$ we have by convergence of $(x_n)\rightarrow x$ that there exists a $N\in\mathbb{N}$ such that $|x-x_n|<\epsilon$ for all $n\ge N$. However, if we now choose $\epsilon<\alpha$, this implies that $x_n-k = -(x-x_n) + (x-k) > -\epsilon + \alpha > 0$ for all $n\ge N$, which is a contradiction to all $x_n\le k$.