If $x_n\leq y_n$ then $\lim x_n\leq \lim y_n$ [duplicate]

real-analysis sequences-and-series limits inequality

Solution 1:

A proof without contradiction:

Suppose that $x_n \leq y_n$ for all $n$. For any $\epsilon > 0$, there exists an $N$ such that $n>N$ implies that $|x_n - x| \leq \epsilon$ and $|y_n - y| \leq \epsilon$. We then note that $$ y - x = (y - y_n) + (y_n - x_n) + (x_n - x) \geq\\ -|y-y_n| + (y_n - x_n) -|x_n - x| >\\ (y_n - x_n) - 2 \epsilon \geq \\ -2 \epsilon $$ So, we have $y - x > -2 \epsilon$ for every $\epsilon > 0$. It follows that $y - x \geq 0$, as desired.

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