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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?

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  • $\begingroup$ 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. $\endgroup$ – Peter Foreman Apr 17 at 0:17
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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 $\forall i \ge k'$,

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

contradicting the limit definition.

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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 $M<x_n <L+(M-L)=M$ which is a contradiction.

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