# Proof

### 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