Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 21

Use the precise definition of a limit to prove the statement $\lim \limits_{x \to 2} (x^2 - 3x) = -2$

From the definition of the limit

$\qquad$ if $0 < | x - a | < \delta $ then $|f(x) - L | < \varepsilon $

$\qquad$ if $0 < | x - 2 | < \delta $ then $|(x^2 - 3x) - (-2)| < \varepsilon$

But,

$\qquad$ $|(x^2 - 3x) - (-2)| = |x^2 - 3x + 2|$

To associate $|x^2 - 3x + 2 |$ to $|x - 2|$ we can factor and rewrite $|x^2 - 3x + 2|$ to $|(x - 2)(x - 1)|$ to obtain from the definition

$\qquad$ if $0 < |x - 2| < \delta$ then $|(x - 2)(x - 1)| < \varepsilon$

We must find a positive constant $c$ such that $|x - 1 |< c$, so $|x - 2| |x - 1| < c |x - 2|$

From the definition, we obtain

$\qquad$ $\displaystyle c |x - 2| < \varepsilon$

$\qquad$ $|x - 2| < \displaystyle \frac{\varepsilon}{c}$

Again from the definition, we obtain

$\qquad$ $ \displaystyle \delta = \frac{\varepsilon}{c}$

Since we are interested only in values of $x$ that are close to 2, we assume that $x$ is within a distance 1 from 2, that is, $-1 < |x - 2| < 1$

Then $1 < x < 3$, so $0 < x - 1 < 2$

Thus, we have $|x - 1| < 2$ and from there we obtain the value of $c = 2$

But we have two restrictions on $|x - 2|$, namely

$\qquad$ $|x - 2| < 1$ and $|x - 2| < \displaystyle \frac{\varepsilon}{c} = \displaystyle \frac{\varepsilon}{2}$

Therefore, in order for both inequalities to satisfied, we take $\delta$ to be smaller to $1$ and $\displaystyle
\frac{\varepsilon}{2}$. The notation for this is $\delta =$ min $\displaystyle \left\{1, \frac{\varepsilon}{2} \right\}$