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How to prove that sequence $x_n$ has a smallest number (comprehensively, preferably by using the definition of the limit of a sequence), when:

$$x_1=0$$ $$\lim_{n\to\infty}x_n=7$$

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    $\begingroup$ Do you mean $\lim_{n \to \infty} x_n = 7$? $\endgroup$ – Omnomnomnom Jan 25 '17 at 7:03
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Hint: Using the definition of convergence, we can state that there exists an $N$ such that $n>N$ implies that $x_n > 0$ (or if you prefer: $n > N$ implies that $|x_n - 7| < 7$).

One of the elements $x_1,x_2,\dots,x_N$ must be the minimum. Why?

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