Continuity of a function

dwsmith

Well-known member
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)?

Plato

Well-known member
MHB Math Helper
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*}.

dwsmith

Well-known member
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?

Plato

Well-known member
MHB Math Helper
Do I have to consider for $\delta > 1$ next too?

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