Question

A function f(x), a point c, the limit of f(x) as x approaches c, and a positive number ε are given. Find a number δ > 0 such that for all x, 0 < |x - c| < δ ⇒ |f(x) - l| < ε. F(x) = - 2x - 6, l = - 12, c = 3, and ε = 0.01

Answer 1

The limit of this function is obviously -12, because the function is actually continuous and we can directly compute it:

$f(3) = -2\cdot 3 - 6 = -6-6 = -12$

So we have $l = -12$. Now, since we want the function and the target to be far no more than 0.01, we have

$|f(x) - l| < 0.01 \iff |-2x-6 - (-12)| < 0.01 \iff |-2x+6|<0.01 \iff |-2(x-3)| < 0.01 \iff 2|x-3|<0.01 \iff |x-3|<\dfrac{0.01}{2} = 0.005$

So, we came to the following conclusion: in order for $f(x)$ and the limit $l$ to be far no more than 0.01, we need the variable x and its target 3 to be far no more than 0.005. In other words, we're claiming that

$0 < |x-3| < 0.005 \implies |(-2x-6) - (-12)| < 0.01$