A question on the Epsilon-Delta definition

Question

I'm starting to study calculus and I stumbled upon the Epsilon-Delta limit definition. It is apparently used as a foundation for the most basic theorems in calculus. So I was trying to use the definition to disprove (by contradiction) that the limit of $f(x) = \frac{x^2-4}{x-2}$ as $x$ approaches $2$ is $5$.

I first used the definition to reach the statements:

$0 \lt \vert x−2 \vert \lt \delta$

$\vert x-3 \vert \lt \varepsilon$ (Already factored and subtracted)

Now I have no idea what to do. If I gave the correct answer, $4$, then it would follow that $\vert x-2 \vert \lt \varepsilon$ and so I could've just defined delta to be equal to epsilon and it would suffice... what am I missing?

Answer 1

For an epsilon-delta limit statement to be true, you need to prove $\forall \varepsilon \gt 0 \exists \delta$ such that a particular statement is true. That means to prove the statement is false, you must prove $\exists \varepsilon \gt 0 \forall \delta$ the same statement is false.

In this case, the statement is that whenever $0 \lt \vert x-2 \vert \lt \delta$ we have that $\vert f(x)-5 \vert \lt \varepsilon$. So choose a specific $\varepsilon$; say, $\varepsilon = 0.5$. You can see pretty easily that there is no $\delta$ that forces $\vert f(x)-5 \vert \lt \frac 12$ whenever $0 \lt \vert x-2 \vert \lt \delta$ because whenever $\vert x-2 \vert \lt \min (\delta, \frac 14)$, the inequality $\vert f(x)-5 \vert \lt \frac 12$ is false.

Answer 2

To get a feel for this, you can graph the function $f(x)$ and (tentatively) define $f(2) = 5$. After this you can inspect the graph and then (with great confidence) set $\varepsilon = 1$. Now for any $\delta \gt 0$ select the the smallest positive integer $n$ such that

$$ \frac{1}{n} \lt \delta$$

Then set $x_0 = 2 - \frac{1}{n}$ and observe that

$$0 \lt \vert x_0−2 \vert \lt \delta \quad \land \quad \vert f(x_0)-5 \vert = \vert 4 - \frac{1}{n}-5 \vert = 1 + \frac 1n$$