How can I prove the continuity of $f$ at $x = 1$?

[Dustinsfl]

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)?

 

[Plato2]

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)?

Start with $0<\delta <1$
$$\begin{align*}|x-1|&< 1\\ 0<x &< 2 \\4< x+4 &<6\\|x+4|&< 6\end{align*}$$.

 

[Dustinsfl]

Start with $0<\delta <1$
$$\begin{align*}|x-1|&< 1\\ 0<x &< 2 \\4< x+4 &<6\\|x+4|&< 6\end{align*}$$.

Do I have to consider for $\delta > 1$ next too?

 

[Plato2]

Do I have to consider for $\delta > 1$ next too?

NO! You let $\delta = \min \left\{ {1,\frac{\varepsilon }{6}} \right\}$