Question

Let f(x)=x2−12x. To prove that limx→6f(x)=−36, we proceed as follows. Given any ϵ>0, we need to find a number δ>0 such that if 0<|x−6|<δ, then |(x2−12x)−(−36)|<ϵ. What is the (largest) choice of δ that is certain to work? (Your answer will involve ϵ. When entering your answer, type e in place of ϵ.)

Answer 1

Answer:

Take $\delta = \sqrt{\epsilon}$

Step-by-step explanation:

To begin notice that

$\big| x^2 -12x-(-36) \big| = \big| x^2 -12x+36 \big| = \big| (x-6)^2 \big|\\= \big|x-6\big|*\big|x-6 \big| \leq \delta*\delta = \delta^2$

If we chose the following delta

$\delta = \sqrt{\epsilon}$

that solves our problem !