Does anyone have a quick way to show that if $f(x)$ tends to a limit A as $x$ tends to $0$, and $f(x) \geq 0$ then $A \geq 0$?



$|f(x)-A|<\epsilon \Rightarrow A-\epsilon < f(x) < A+\epsilon$

If $A<0$ then choose $\epsilon$ less than $|A|$ which forces $f(x)<0$.


Let if possible $A<0$

Then $-A>0$

As $f(x)\to A$ as $x\to 0$ so $\forall \epsilon>0 \exists \delta>0$ such that $|f(x)-A|<\epsilon$ for all $x$ with$|x|<\delta$

Take $\epsilon=-A/2$

Then we have $|f(x)-A|<-A/2\Rightarrow 3A/2<f(x)<A/2<0$ for all $|x|<\delta_A$ This is a contradiction.So $A\ge 0$


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