# proving that delta exists to a limit

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.

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 $-1 < f(x) < 11$.
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 $|x-1|<\delta$.
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.