# Sequence\Limits Proof

Let $$L = \lim_{k \rightarrow \infty}\limits x_k$$. If $$(x_k)_{k=0}^\infty$$ is increasing, then $$x_k \le L$$ for all $$k \ge 0$$

Could anybody push me in the right direction? I've stared at this one for a while and I'm not sure how to get this proof started. Is induction the way to go?

• Proof by contradiction. If we had $x_k\gt L$ for some $k\in\mathbb{N}$ then the value of the sequence must decrease in order for the limit to be $L$. This is a contradiction of the sequence being increasing. – Peter Foreman Apr 17 at 0:17

If we had $$x_k > L$$ for some $$k\in \mathbb N$$ and $$x_k$$ is increasing, then $$\forall i > k$$

\begin{align*} x_i &> x_k\\ x_i - L &> x_k - L \end{align*}

where $$x_k-L > 0$$.

Then there is no $$k'$$ such that $$\forall i \ge k'$$,

$$|x_i -L| < x_k - L$$

Suppose $$x_k >L$$ for some $$k$$. Pick any number $$M$$ in $$(L, x_k)$$. Now $$x_n \geq x_k >M$$ for all $$n \geq k$$. But $$|x_n -L| < M-L$$ for $$n$$ sufficiently large. For such $$n$$ we get $$M which is a contradiction.