Proved by defining the boundary (delta and epsilon)

Question

I would be happy if anyone could give me a detailed answer. I was unable to express Delta using Epsilon.

My thought was (I will use keyboard keys) |x-a| = |x -(- 7)| = |x + 7| <delta | 3x +20 -1 | = |x + 7 + 2x + 12| < |x + 7| + 2| x + 6| I now have an expression | x + 7 | Which is smaller than Delta, but how do I reduce 2|x+ 6|

\begin{equation*} \lim_{x \rightarrow -7} |3x + 20|=1 \end{equation*}

Answer 1

Let $\epsilon>0$. By the reverse triagle inequality we have:

$||3x+20|-1|=||3x+20|-|-1||\leq |(3x+20)-(-1)|=|3x+21|=3|x+7|$

So now let $\delta=\frac{\epsilon}{3}$. If $|x-(-7)|=|x+7|<\delta$ then we get:

$||3x+20|-1|\leq 3|x+7|< 3\delta=3\frac{\epsilon}{3}=\epsilon$