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Proof

(related to Proposition: Preservation of Inequalities for Limits of Functions)

Context

• Let $D\subseteq\mathbb R$ be a subset of real numbers.
• Let $a\in\mathbb R$ be a real number.
• Let $f:D\to\mathbb R$ be a function with the limit $\lim_{x\to a}f=L.$

Case 1: Let $f(x) > 0$ for all $x\in D.$

 * Hypothesis:
    * Assume `$L < 0.$`
 * Implications
    * By the [definition of limit][bookofproofs$6683], for `$\epsilon=-L$` there is a `$\delta > 0$` such that for all `$x\in D$` satisfying `$0 < |x-a| < \delta,$` it follows that `$|f(x)-L| < - L.$`
    * For these values of `$x$` we have `$-L > |f(x)-L| = f(x) - L,$` implifying that `$f(x) > 0$` and `$L < 0.$`
    * This creates the [contradiction][bookofproofs$744] to `$f(x) < 0.$` 
 * Conclusion
    * It must be that `$L\ge 0.$`

Case 2: Let $f(x) < 0$ for all $x\in D.$

 * Hypothesis:
    * Assume `$L > 0.$`
 * Implications
    * By the [definition of limit][bookofproofs$6683], for `$\epsilon=L$` there is a `$\delta > 0$` such that for all `$x\in D$` satisfying `$0 < |x-a| < \delta,$` it follows that `$|f(x)-L| < L.$`
    * For these values of `$x$` we have `$L > |f(x)-L| = -f(x) + L,$` implifying that `$f(x) < 0$` and `$L > 0.$`
    * This creates the [contradiction][bookofproofs$744] to `$f(x) > 0.$` 
 * Conclusion
    * It must be that `$L\le 0.$`<div class='qed'>&#8718;</div>

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References

Bibliography

1. Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016