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- #1

Given $\varepsilon > 0$.

There exist a $\delta > 0$ such that $|x - c| < \delta$ whenever $|f(x) - f(c)| < \varepsilon$.

From the statement of the $\varepsilon-\delta$ definition, we have that

$$

|x - 1| < \delta\quad\text{and}\quad |f(x) - f(1)| < \varepsilon.

$$

Let's look at $|f(x) - f(1)| = |x^2 + 3x - 4| < \varepsilon$.

Then

\begin{alignat*}{3}

|x^2 + 3x - 4| & = & |(x - 1)(x + 4)|\\

& = & (x + 4)|x - 1|

\end{alignat*}

What can I do about (x + 4)?