In the last meal [calculus]The clarification of ε-δ definition, we knew the actual meaning of \(\varepsilon -\delta\) definition. This time we are going to use it. In this dish, I will show you the basic problem in \(\varepsilon -\delta\) definition. Hope to make you understand clearly!

## The condition of「there exists δ >0」

- to find \(\delta >0\), implicityly “to prove \(\displaystyle{\lim_{x \to c}f(x)=L}\)”That is, the definition not hold yet, so we have to figure $latex \delta >0 out.
- we get \(\delta >0\), implicityly “we know \(\displaystyle{\lim_{x \to c}f(x)=L}\)”In this condition, the definition have already held（So we get \(\delta >0\)）

## Let’s see some problem

Before that, I suggest that you should be familiar with \(\varepsilon -\delta\) definition (or you can put it aside you for comparison), so that you know the meaning of this topic. Let’s start to solve the problem!

##### Basic type

This type of question can be said to be the must-do question when you knew the \(\varepsilon -\delta\) definition first time:

Show that \(\displaystyle{\lim_{x \to 2}3x=6}\)

proof:

At first, we have to give an \(\varepsilon\): Let \(\varepsilon>0\)

And then we are going to find \(x\) that satisfied the inequality \(\left | f(x)-L \right | < \varepsilon\).

In this case, \(\left | 3x-6 \right | < \varepsilon\)

So the next step is to use the property we known (\(\left | 3x-6 \right | < \varepsilon\)), and think how to get \(x-2\) from this inequality.

Simplify the inequality:

\(\left | 3x-6 \right | < \varepsilon\)

\(\Rightarrow 3\left | x-2 \right | < \varepsilon\)

\(\Rightarrow \left | x-2 \right | < \displaystyle{\frac{\varepsilon}{3}}\)

If you know the definition well, you could easily find that \(\displaystyle{\frac{\varepsilon}{3}}\) represents the largest interval I could get corresponding to x with the situation of this question (Hope that \(f(x)=3x\) is closed to 6).

Now we have the interval, I can call this thing, \(\displaystyle{\frac{\varepsilon}{3}}\) as \(\delta\).

Then finish the conclusion by math language:

Take \(\delta = \displaystyle{\frac{\varepsilon}{3}}\).

Then \(\forall x \; in \; 0 < \left | x-2 \right | < \delta, \; \left | 3x-6 \right | < \varepsilon \)

(We can see the statement of \(\varepsilon -\delta\) definition in this part.)

Finally, remember mention the thing you want to prove at first:

Therefore \(\displaystyle{\lim_{x \to 2}3x=6}\)

The complete proof looks like this:

Let \(\varepsilon>0\)

\(\left | 3x-6 \right | < \varepsilon\)

\(\Rightarrow 3\left | x-2 \right | < \varepsilon\)

\(\Rightarrow \left | x-2 \right | < \displaystyle{\frac{\varepsilon}{3}}\)

Take \(\delta = \displaystyle{\frac{\varepsilon}{3}}\).

Then \(\forall x \; in \; 0 < \left | x-2 \right | < \delta, \; \left | 3x-6 \right | < \varepsilon \)

Therefore \(\displaystyle{\lim_{x \to 2}3x=6}\)

##### Some different types

The type of questions are basically the same. There are just some changes in the process.

Show that (\displaystyle{\lim_{x \to 2}x^2=4})

proof:

Like previous question, we have to give \(\varepsilon\)：Let \(\varepsilon>0\)

Then we have to find \(x\) that satisfied the inequlity \(\left | x^2-4 \right | < \varepsilon\). We want to discuss the behavior of x closed to 2, so we use the condition we know (\(\left | x^2-4 \right | < \varepsilon\)) to find the range of \(x-2\).

Simplify the inequality:

\(\left | x^2 -4 \right |\)

\(=\left | (x-2) \right | \left | (x+2) \right | \)

\(=\left | x-2 \right | \left |x+2 \right | < \varepsilon\)

We also get an interval of \(x-2\) but it seems a little different:

\(\left |x-2 \right | < \displaystyle{\frac{\varepsilon}{\left |x+2 \right |}}\)

The different is that the equation on the right-hand side is a function, so we have to deal with the \(x\) .

In terms of the definition, it doesn’t matter that how you choose the range, but can’t be far away from the point 2. I can choose any number, 1 for example. And then we use the range we choose to estimate \(x+2\) (remember this is our target).

Assume \(\left |x-2 \right | < 1\) （notice that I choose it by myself）

Then \(-1< x-2 < 1\)

\(\Rightarrow 1< x < 3\)

\(\Rightarrow 3< x+2 < 5\)

\( \Rightarrow \left | x+2 \right | < 5\)

So we can turn back to the place we stuck and continue simplifying the equation.

\(\left | x^2 -4 \right |\)

\(=\left | x-2 \right | \left |x+2 \right | <5\left | x-2 \right |\)（ because \(\left | x+2 \right | < 5\) ）

\(< \varepsilon\)

\(\Rightarrow \left | x-2 \right |< \displaystyle{\frac{\varepsilon}{5}}\)

Now we successfully get back to the definition. However there is something to notice:

We choose a range of \(x\) by ourselves, right? But we can’t be sure if the number we choose would smaller than \(\displaystyle{\frac{\varepsilon}{5}}\) (after all, what we discuss is the approach to \(x\). we can’t instead take a number that not closed enough). As a result, we need to choose the smaller one to avoid this situation happen.

Take \(\delta = min(1,\displaystyle{\frac{\varepsilon}{5}})\)

Then \(\forall x \; in \; 0 < \left | x-2 \right | < \delta,\;\left | x^2-4 \right | < \varepsilon\)

Therefore \(\displaystyle{\lim_{x \to 2}x^2=4}\)

The complete proof looks like this:

Let \(\varepsilon>0\)

\(\left | x^2 -4 \right | \)

\(=\left | (x-2) \right | \left | (x+2) \right | \)

\(=\left | x-2 \right | \left |x+2 \right | \)

\(< \varepsilon\)

\(\Rightarrow \left |x-2 \right | < \displaystyle{\frac{\varepsilon}{\left |x+2 \right |}}\)

Assume \(\left |x-2 \right | < 1\)

Then \(-1< x-2 < 1\)

\(\Rightarrow 1< x < 3\)

\(\Rightarrow 3< x+2 < 5\)

\( \Rightarrow \left | x+2 \right | < 5\)

\(\left | x^2 -4 \right | =\left | x-2 \right | \left |x+2 \right | \)

\(<5\left | x-2 \right |\)

\(< \varepsilon\)

\(\Rightarrow \left | x-2 \right |< \displaystyle{\frac{\varepsilon}{5}}\)

Take \(\delta = min(1,\displaystyle{\frac{\varepsilon}{5}})\)

Then \(\forall x \; in \; 0 < \left | x-2 \right | < \delta,\;\left | x^2-4 \right | < \varepsilon\)

Therefore \(\displaystyle{\lim_{x \to 2}x^2=4}\)

##### Add some ingredients

♦ [calculus]The clarification of ε-δ definition

♦ 清大數學系開放式課程-微積分一（高淑蓉）

##### Seasoned with Chinese

◊ limit – 極限

◊ calculus – 微積分

◊ definition – 定義

Pictures credit:

article feature picture and the graph of this article are made by Alex.