The function is: $\ f(x) = \begin{cases} \frac{7}{100}x^2-\frac{3}{5}, & x\le 1 \\\\ \frac{100x^2-137x+37}{100(x-1)}, & x > 1 \end{cases}$
For $x_0=1$ find an $\varepsilon$>0 such that for every $\delta$>0 is true that: for $|x-1|$<$\delta$ $ \Rightarrow$ $|f(x)-f(1)|>$ $\varepsilon$.
My questions are 1) Which of the two possible options for $f(x)$ do we use in the $|f(x)-f(1)|>$ $\varepsilon$ and why?? $\\$ and 2) How do we find epsilon?
Just for a reference the answer given is $\varepsilon$=$\frac{3}{100}$