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The question is :

Assume $\lim_{x\rightarrow1}f(x)=5$. Prove that there exists $\delta>0$ s.t. for every $x$ that sustains the condition $|x-1|<\delta$, we know that $f(x)>-1$.

I know that because it is known that there is a limit I can choose an epsilon, so if I'll say epsilon = $1$ then $4 < f(x) < 6$. I also know that $1 - \delta < x < 1 + \delta$ yet I am not sure how that helps me.

Thanks in advance !

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By what is given there is some $\delta > 0$ such that $|x-1| < \delta$ implies $|f(x) - 5| < 6$, implying that $-6 < f(x) - 5 < 6$, implying that $-1 < f(x) < 11$.

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  • $\begingroup$ Actually is not by continuity but by the existence of the limit. $\endgroup$ – EA304GT Nov 15 '15 at 15:14
  • $\begingroup$ @EA304GT Thanks; did not pay attention to that. :) $\endgroup$ – Megadeth Nov 15 '15 at 15:17
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I don't really see the interest of the question. But, precisely as you said, there is $\delta>0$ s.t. $$|x-1|<\delta\implies |f(x)-5|<1$$ and thus, if $|x-1|<\delta$ $$4<f(x)<6$$ in particular, $f(x)>-1$ if $|x-1|<\delta$.

I wanted to put it as a comment, but it has to many characters. That's why I posted as an answer, but it's not really an answer.

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We know that for each $\epsilon>0$ there is a $\delta >0$ such that $|x-1|<\delta$ implies $|f(x)-5|<\epsilon$, that is $-\epsilon<f(x)-5<\epsilon$. Thus, $$5-\epsilon<f(x)<5+\epsilon$$ We need $\epsilon$ such that $5-\epsilon =-1$, hence, $\epsilon=6$ will do.

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