Working with the (ε,δ) definition of limits to find the δ of

Question

I'm introducing myself to the $(\varepsilon, \delta)$ definition of limits, and I'm encountering a few issues.

When proving the $\lim_{x \to c}f(x) = L$ $$ \forall \varepsilon > 0, \ \exists \delta = \delta(\varepsilon) > 0 : 0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon $$

When considering $\lim_{x \to 2}(2x - 5) = -1$

Let $\forall \varepsilon > 0$

Choose $\delta = \dfrac{\varepsilon}{2}$

Assume $0 < |x - 2| < \delta$

Then,

$$ |2x - 5 - (-1)| < \varepsilon, $$

$$ \\|2x - 4| < \varepsilon \\2|x - 2| < \varepsilon \\|x - 2| < \delta \\2|x - 2| < 2\delta \\ \therefore \delta = \dfrac{\varepsilon}{2} $$

Forgive my mistakes, I'm still quite new to this. I believe that my proof is mostly accurate (do correct me if I'm wrong, please).

My biggest issue comes with solving this other problem:

I am to suppose $|f(x)-7| < 0.2$ whenever $0 < x < 7$.

Find all values of $\delta > 0$ such that $|f(x) - 7| < 0.2$ whenever $0 < |x-2| < \delta$.

I've not got a good idea of how to approach this with an arbitrary $f(x)$.

Answer 1

We want to find all $\delta > 0$ such that $ 0 < |x-2| < \delta $ implies $ 0 < x < 7$. We distinguish two cases:

1. $x \geq 2$: Then we have $0 < x - 2 < \delta \; \Leftrightarrow \; 2 < x < \delta + 2$. From the condition $ x < 7$ we see that $\delta \leq 5$ must be satisfied.
2. $ x < 2$: Then we get $ 0 < 2 - x < \delta \; \Leftrightarrow \; -2 < -x < \delta - 2 \; \Leftrightarrow \; 2 - \delta < x < 2$. For $\delta \leq 2$ the condition $ 0 < x$ is satisfied.

So we have two conditions on $\delta$ : $\delta \leq 2$ and $\delta \leq 5$: Clearly, we can reduce this to $\delta \leq 2$. Hence the set of all $\delta > 0 $ such that $0 < |x-2| < \delta$ implies $0 < x < 7$ is exactly the set $(0,2]$.