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Ignoring the fact that there is a much simpler proof using product of limits, is this $\epsilon$ proof solid for the question in the title?


Since $x_n \rightarrow 5$ as $n \rightarrow \infty$ then since $\frac{\epsilon}{11} > 0, \exists N_1$ s.t. $\forall n \geq N_1$ we have $|x_n -5| < \frac{\epsilon}{11}$. Likewise, since $\epsilon = 1 >0, \exists N_2$ s.t. $\forall n \geq N_2$ we have $|x_n -5| < 1 \implies x_n < 6$.

Then let $N = \max\{N_1, N_2\}$ then $\forall n \geq N$ we get:

$|x_n^2 - 25| = |x_n -5||x_n + 5| < 11|x_n - 5| < \frac{11\epsilon}{11} = \epsilon $

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    $\begingroup$ Seems pretty solid. $\endgroup$ – Jacky Chong Apr 22 '17 at 21:46
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You actually need to assert not only that $x_n<6$ but also that $x_n>-4$ in order to claim that $|x_n+5|<11$. But this is a detail many teachers wouldn't worry about.

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