Saturday, May 26, 2018

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

Show that the statement lim 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

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