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=
, so more than ), nor any of the first five I wrote.
It will not be the fiftieth element, because it is equal
, and its distance from 2 is 0.02, which is still greater than the given value .
However, the hundredth element satisfies the given condition, because
, its distance from 2 is equal to 0.01, which is less than 0.017. However, there are elements with numbers less than 100 that also meet this condition. Our task is to find the limit position from which the distance of the elements of the sequence from 2 is less than 0.017. Since the distance of the fiftieth element does not meet the condition but the hundredth one does, it seems that it will be some element between the fiftieth and the hundredth… How to find it exactly?
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…
from this definition it is the general term of the sequence, in our example it is equal to .
we have it predetermined , 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
. This expression is always positive because n stands for the position of the sequence elements and we take numbers such as: 1, 2, 3,… . So we have the absolute value of a positive number. We can therefore leave it out, because the absolute value of a positive number is always positive:
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
. We need to calculate from which positions the sequence’s elements distances from the limit will be smaller than this value.
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
is always positive, so we can write:
We can multiply both sides by
, because is always greater than zero…
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 there is a position ‘N’ of element in the sequence from which subsequent elements would satisfy the inequality:
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
(because it is always positive).
We divide both sides by
(we can, because in the definition of the limit is always positive):
And now: n are the positions of the elements in the sequence. They are getting bigger and bigger. For any we determine the left side of the inequality, we will get some constant number there. Even if it would be very large, however, as ‘n’ grows and grows, eventually the right side of the inequality will “jump over” the left side and from some n onward the inequality will be satisfied, regardless of the choice. .
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 
, i.e. the absolute value at the bottom is calculated from a positive number.
We multiply both sides by the denominator (for the same reasons we left out the absolute value):
This time we cannot divide by 
like the last time, because for certain values the expression is positive, and for some the expression is negative. If we agreed that , for example, is very tiny (we can, because it could be anything), expression will be negative and after dividing both sides by: (remember to change the sign of the inequality after multiplying both sides by a negative value!):
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) –>