## Limit of a sequence Lecture 2

### Topic: Determining the limit of a sequence by definition

#### Summary

In the article I will show how in practice you can “feel” the definition of the limit of a sequence using specific examples. Let’s recall the definition of the limit of a sequence:

**Definition of a sequence limit**

The number g is called the limit of a sequence if and only if:

\underset{\varepsilon >0}{\mathop{\forall }}\,\underset{N}{\mathop{\exists }}\,\underset{n>N}{\mathop{\forall }}\,\left| {{a}_{n}}-g \right|<\varepsilonThe idea is that for any small, chosen in advance, epsilon ( \varepsilon ), we can find the position in the sequance of the element ( N ) that all elements with consecutive positions greater than N ( n> N ) would have a distance from the boundary g smaller than epsilon ( \left| {{a} _{n} }-g \right|< \varepsilon ) – no matter how small this fixed epsilon is.

Difficult? Well, of course it’s difficult. It takes practice to understand. We will see how it works with below examples.

#### Example 1 for calculating which elements of a sequence satisfies the inequality for a given epsilon

*It’s given a sequence with a general term and border . For epsilon equal – which elements of the sequence satisfy the inequality ?*

In the problem, we set the epsilon to 0.017. We need to calculate from which element’s number the distances of the sequence terms from the limit will be smaller than 0.017. Get out your calculator, OK?

First, very slowly, we’ll figure out what’s going on…

Our sequence written element by element would look like this:

I obtain subsequent elements by substituting specific positions for n. Putting it in order, I will have:

That is:

The subsequent elements of the sequence will be closer and closer to the number 2. Our task is to find the position of the sequence element from which the distances from number 2 will be smaller than the given ones .

It certainly won’t be the first element. It is equal to 3. Its distance from the border (g=2) is equal to 1! This is much more than requested .

It will not be the second element either (distance=), nor any of the first five I wrote.

It will not be the fiftieth element, because it is equal.

However, the hundredth element satisfies the given condition, because

Let’s recall the definition…

\underset{\varepsilon >0}{\mathop{\forall }}\,\underset{N}{\mathop{\exists }}\,\underset{n>N}{\mathop{\forall }}\,\left| {{a}_{n}}-g \right|<\varepsilonAnd the inequality that comes with it…

.

*, and the limit of g is 2. We put all this into the inequality and we have:*

We reduce the absolute value in the middle to a common denominator:

We subtract and we have:

That is:

Now it’s a little harder. On the left side of the inequality in the absolute value we have

We multiply both sides by n, again we must remember that ‘n’ is always positive and it will not change the sign of the inequality. After multiplying both sides, we get:

Now we divide both sides by 0.017 (we use a calculator and take some reasonable approximation).

Looking at this inequality and remembering that ‘n’ stands for the position of the element in the sequence – what will be the answer?

**Answer**

All elements of the sequence numbered from 59 upwards satisfy the inequality. In this sequence, the distance of the 58th element from 2 is not smaller than the given epsilon, but the 59th, 60th, 61st,… and each subsequent one is.

#### Example 2 for calculating which elements of the sequence satisfies the inequality for a given epsilon

*Given is a sequence with a general term and limit . For epsilon equal – which elements of the sequence satisfy the inequality ?*

This example is very similar to the previous one. We have epsilon set to

Let’s break down the sequence:

That is…

The elements of the sequence become smaller and smaller, approaching zero. You need to determine which ones have this distance smaller than*.*

It won’t be that easy again, you can see that it won’t be any of the first five elements. Let’s get straight to the point…

We insert the appropriate values into the inequality from definition:

That is…

We can skip the absolute value again because

We can multiply both sides by

We multiply both sides by 10235 (to make it easier) and we have:

We move 27 to the left:

We divide both sides by 27 (calculator) and approximately:

‘n’ can only be positive (it makes life easier, right?) so we can write:

So approximately:

So we have:

**Answer**

All elements of the sequence starting from the 20th position satisfy the inequality.

#### Example 1 for calculating the limit of a sequence from definition only

Let’s do another task in which we will have to dig into the definition of the limit of the sequence:

*Show by definition that the limit of a sequence is is the number .*

Using the definition, it is necessary to show that for any arbitrarily small

And in our specific example, we need to show that for any small fixed epsilon, for all positions n starting from some position ‘N’, the inequality will be satisfied:

We will do this by determining ‘n’ from the above inequality. First, let’s reduce it to a common denominator…

After some cleaning:

We can break down this absolute value on the left – for example – like this:

That is:

We can multiply both sides by

We divide both sides by

And now: n are the positions of the elements in the sequence. They are getting bigger and bigger. For any

So we showed by definition what we were supposed to show.

Finally, let’s do an example the other way around…

#### Example 2 for calculating the limit of a sequence by definition

*Check from the definition whether the limit of the sequence , where is greater than 1, is a number .*

Let’s pay attention to a nuance. In the previous example, the text of the task was “Prove that…” – that is, it was known in advance that that given number was indeed the limit of the sequence and it only needed to be proven. Here we have “Check if…” – so perhaps our number will not be the limit of the sequence at all.

We start with inequality from the definition…

To which we substitute appropriate values…

We try again to determine ‘n’ from the above inequality.

Absolute values can be omitted because – pay attention to the text of the problem – it is said that

We multiply both sides by the denominator (for the same reasons we left out the absolute value):

This time we cannot divide by

So we have (due to the reversed inequality sign) a completely opposite situation than in the previous example. On the left there is a fixed number. On the right, the positions’s ‘n’ are getting bigger and bigger. It can be stated that from certain ‘n’s the inequality will NOT be fulfilled (and it was supposed to be fulfilled for any epsilon).

Therefore, the number from the problem is not the limit of the sequence.

Click to recall the definition of the limit of a sequence (previous Lecture)< —

Click to learn more about undefined expressions (next Lecture) –>