[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).

calculus-limit-epsilon-delta-definiton-example-01

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

calculus-limit-epsilon-delta-definiton-example-02

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


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.

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