Consider , where f(x) = $a x - $b. Show that |f(x) - $c| < $a δ if |x -$x| < δ.
Find δ such that |f(x) - $c| < $d if |x - $x| < δ.
Prove rigorously that .
To show that |f(x) - $c| < $a δ if |x - $x| < δ, we substitute f(x) = $a x - $b into the expression |f(x) - $c| and algebraically simplify.
|f(x) - $c| = |($a x - $b) - $c|
= |$a x - $bplusc|
= nc1ItEz0kR4= |x - $x|
Now use the fact that |x - $x| < δ to show |f(x) - $c| < $a δ.
|f(x) - $c| = nc1ItEz0kR4= |x - $x|
< nc1ItEz0kR4= δ
Find a δ such that |f(x) - $c| < $d if |x - $x| < δ. In Step 1 we showed that |f(x) - $c| < $a δ if |x - $x| < δ.
Therefore we can set $a δ = $d and solve for δ. This will give us the maximum δ that satisfies the conditions. (Round your answer to six decimal places.)
$a δ = $d
δ = qjqZz1N+poo=
Prove rigorously that .
Recall the formal definition of a limit. Suppose that f(x) is defined for all x in an open interval containing c (but not necessarily at x = c). Then if for all ε > 0, there exists δ > 0 such that |f(x) - L| < ε if 0 < |x - c| < δ.
According to this definition, to prove that , we must show that for all ε > 0, there exists δ > 0 such that |f(x) - SFgqQUkJGdg=| < ε if 0 < |x - i0sYGeK9R9c=| < δ.
Algebraically manipulate |f(x) - $c| so that we get a factor of |x - $x|. Then solve for δ in terms of ε.
|f(x) - $c| = $a |x - $x| < ε
We see the gap between f(x) and $c is $a times the gap between x and $x, which we wish to call δ. Since we also wish to keep |f(x) − $c| < ε, we need to choose δ in terms of ε so that |x − $x| < δ and this equation holds. We do this by choosing the following value of δ.
δ = ε/nc1ItEz0kR4=
This value of δ satisfies the formal definition of limit and proves .