Prove that for every $\varepsilon > 0$ there exists $\delta > 0$ such that if $|x − 3| < \delta$, then $|x^2 − 5x + 6| < \varepsilon$.

Question

I know that I will need to find a suitable $\delta$, but I'm having trouble starting. Any help is appreciated!

Answer 1

• Suppose that $|x-3| \lt \delta$.

• Note that the polynomial factors as $(x-2)(x-3)$.

• Apply the rule : $|ab| = |a|\times|b|$.

• The goal can be rewritten as : $|x-3|\times |x-2| \lt\epsilon$.

• The question is : if the first factor is less than $\delta$ ( this is your hypothesis) , how does this affect the value of the second factor , and what delta can I choose to make the whole product less than $\epsilon$.

• We have to express the value of the product in terms of delta, in order to find a saitable $\delta$ expressed in terms of $\epsilon$.

• Let's first see how the value of the whole product can be expressed in terms of delta

$|x - 3| \lt \delta$

$\rightarrow -\delta\lt x-3\lt\delta$

$\rightarrow -\delta+3\lt x\lt\delta+3$

$\rightarrow -\delta+1\lt x-2\lt\delta+1$

$\rightarrow |x-2|\lt\delta+1$

$\rightarrow |x-3|\times|x-2|\lt\ \delta\times ( \delta+1)$ .

• So in case $|x-3|$ is less than $ \delta$ you are guaranteed that the product ( that is, your original polynomial transformed) is less than $\delta\times ( \delta+1)$.

How can you secure, by playing with the value of delta and expressing it in terms of $\epsilon$ , that your polynomial is also less than $\epsilon$?

For example, if I set $\delta = \epsilon$, that does not work, because it only secures the polynomial to be less than $\delta\times ( \delta+1) = \epsilon(\epsilon+1) $, which is not less than $\epsilon$ ( knowing this number is supposed to be >0)

In the same way setting $\delta = \sqrt{\epsilon}$ does not work.

But we may be on the right track. Are there smaller roots than square roots, that would do?