Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 19

Show that the statement $\lim \limits_{x \to 2} (14 - 5x) = 4$ using the precise definition of a limit.

Based from the definition,

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

$\qquad$ if $0 < | x - 2 | < \delta $ then $|(14-5x)-4| < \varepsilon$

But,

$\qquad$ $|(14-5x)-4| = |14 - 5x -4| = |10 - 5x| = |-5 (x - 2)| = 5 |x - 2|$

So we want

$\qquad$ if $0 < |x - 2| < \delta$ then $5| x - 2 | < \varepsilon$

That is,

$\qquad$ if $0 < | x - 2 | < \delta$ then $|x - 2| < \displaystyle \frac{\varepsilon}{5}$

The statement suggest that we should choose $\displaystyle \delta = \frac{\varepsilon}{5}$.

By proving that the assumed value of $\displaystyle \delta = \frac{\varepsilon}{5}$ will fit the definition.

$\qquad$ if $0 < |x - 2| < \delta$ then,

$\qquad$ $|(14-5x)-4| = |14 - 5x -4| = |10 - 5x| = |-5 (x - 2)| = 5 |x - 2| < 5 \delta = \cancel{5} \left( \frac{\varepsilon}{\cancel{5}} \right) = \varepsilon$

Thus,

$\qquad$ if $0 < | x - 2 | < \delta $ then $|(14-5x)-4 | < \varepsilon$

Therefore, by the precise definition of a limit

$\qquad$ $\lim \limits_{x \to 2} (14-5x) = 4$