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Give a $\varepsilon-\delta$ proof that the function $f$ given by the formula $f(x) = x^2 + 3x - 3$ is continuous at $x = 1$.
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)?
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)?