# [calculus]The example of ε-δ definition problem

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」

There are two conditions to discuss：
1. 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.
2. 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 $$\displaystyle{\lim_{x \to 2}x^2=4}$$

Here I want to tell you is no matter how I set the range (even different size on different side),
it’s okay as long as I don’t set the range far away from the point we focus on.

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 $$\displaystyle{\lim_{x \to 2}x^2=4}$$

##### Seasoned with Chinese

◊ limit – 極限
◊ calculus – 微積分
◊ definition – 定義

Pictures credit: