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Let $\{x_n\}$ be a sequence and $x \in \mathbb{R}$. Suppose for any $e>0$ there is an M such that for $n \geq M, |x_n-x| \leq e$. Show that $\lim x_n = x$.

Proof : since we have that for any $e>0$ there is an M such that for $n \geq M, |x_n-x| \leq e$ it holds that $x_n-x$ it must be that $x_n = x$ or otherwise it wouldnt hold. then since $x_n = x$ for all $x_n$ then $x_n$ is a constant sequence $ \Rightarrow \lim x_n = x$.

Is the proof okay? Please help me.

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    How were you defined $\lim_{n\to\infty}x_n=x$? – 2017-02-06
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    Isn't that just the definition of the expression "$\lim_{n\to\infty} x_n=x$" ? – 2017-02-06
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    By the way, your reasoning is majorly flawed. It is not true, as you claim, that $x_n=x$ eventually. Consider $x_n=1-\frac1n$ and $x=1$ – 2017-02-06
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    "For every $e$ there is an $M$" is drastically different form "The is an $M$ such that for every $e$". You seem to think about the latter when you really have the former. – 2017-02-06

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I suppose that the sequence $(x_n)$ is a sequence in $\mathbb N$ or in $ \mathbb Z$.

If this is so, then your proof is correct, if you choose $e=1/2$. But you should fill your proof with some more details.

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    That would not make it "real-analysis", though. – 2017-02-06
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    Why would that not make it "real-analysis" ?? – 2017-02-06
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    What point is there in assuming that $x_n\in\Bbb Z$ for all $n$, when even the hypothesis suggests $\lim_{n\to\infty}x_n\in\Bbb R$? For the sake of justifying the most obvious inversion of quantifiers known to mankind? – 2017-02-06