# Notes - Analysis I MT22, Limit Inequality Theorem

> Source: https://ollybritton.com/notes/uni/prelims/mt22/analysis-i/notes/limit-inequality-theorem/ · Updated: 2023-01-03 · Tags: notes, analysis, mt22

- [Course - Analysis MT22](https://ollybritton.com/notes/uni/prelims/mt22/analysis-i/)

### Flashcards
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What contradiction do you use to prove that limits preserve weak inequalities, i.e. if $a_n \to L$, $b_n \to M$ and  $a_n \le b_n$ for all $n$ then $L \le M$?::
Assume $L > M$.

^limit-inequality-contradiction
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When proving that limits preserve weak inequalities, i.e. if $a_n \to L$, $b_n \to M$ and  $a_n \le b_n$ for all $n$ then $L \le M$, what value of $\epsilon$ do you take in $|a_n - L| < \epsilon$ and $|b_n - M| < \epsilon$?::
$$
\epsilon = \frac{1}{2}(L - M)
$$

^limit-inequality-epsilon-choice
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<!-- flash-start -->
When proving that limits preserve weak inequalities, i.e. if $a_n \to L$, $b_n \to M$ and  $a_n \le b_n$ for all $n$ then $L \le M$, you use $\epsilon = \frac{1}{2}(L - M)$. What expression are you trying to show true for a contradiction?::
$$
L - \epsilon < a_n \le b_n < M + \epsilon
$$

^limit-inequality-contradiction-expression
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<!-- flash-start -->
When proving that limits preserve weak inequalities, i.e. if $a_n \to L$, $b_n \to M$ and  $a_n \le b_n$ for all $n$ then $L \le M$, you “expand out” $|a_n - L| < \epsilon$ and $|b_n - M| < \epsilon$ into what?::
- $L - \epsilon < a_n < L + \epsilon$
- $M - \epsilon < b_n < M + \epsilon$

^limit-inequality-expand-definitions
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