Kalkulus! LIMITS...
by kamatadu, Aug 7, 2022, 5:21 PM
So! First things first. Forget about all the school Limits you know before going on further 
(JK JK, skul too important for pro pipils like u to forget 
). YUH!! Lets get started.
LIMITS OF A SEQUENCE
Definitions as such are especially hard to understand without examples, so let's go through few examples.
Next problem
.
Now lets see two ways to disprove the limit.
Now for the other one,
To finish with this part, let's have a teaser problem (solution left to the reader to prove )
(JK JK, skul too important for pro pipils like u to forget ). YUH!! Lets get started.LIMITS OF A SEQUENCE
Quote:
The number 
is called the limit of the sequence 
as 
if for any 
, there exists a number 
such that the inequality 
holds true for all 
.
A sequence which has a finite limit is said to be convergent.
A sequence
is said to be infinitely small if 
, and infinitely large if 
.
.
is called the limit of the sequence 
as 
if for any 
, there exists a number 
such that the inequality 
holds true for all 
.A sequence which has a finite limit is said to be convergent.
A sequence
is said to be infinitely small if 
, and infinitely large if 
.
.Definitions as such are especially hard to understand without examples, so let's go through few examples.
Problem wrote:
Problem: Find where the sequence 
converges to.
Solution: First we claim that the sequence
converges to 
.
So, for any, it suffices to find a natural number 
such that for any natural number 
, the inequality 
holds.
Now, lets backtrace for our. The inequality 
is satisfied if
So, if we take
, we are done, and so we conclude that, 
converges to.Solution: First we claim that the sequence
converges to 
.So, for any
, it suffices to find a natural number 
such that for any natural number 
, the inequality 
holds.Now, lets backtrace for our
. The inequality 
is satisfied if
So, if we take
, we are done, and so we conclude that, 
Next problem
.
Problem wrote:
Problem: Given sequence 
. It is given that 
. Find the number of points lying outside the open interval 
.
Solution: The distance from the point to the point 
is equal to 
Outside the interval, there will appear those terms of the sequence for which this distance exceeds 
, i.e., 
whence 
hence, 703 points, namely, 
are found on the outside of the interval.
. It is given that 
. Find the number of points lying outside the open interval 
.Solution: The distance from the point
to the point 
is equal to 
Outside the interval, there will appear those terms of the sequence for which this distance exceeds 
, i.e., 
whence 
hence, 703 points, namely, 
are found on the outside of the interval.Now lets see two ways to disprove the limit.
Problem wrote:
Problem: Prove that 
is not the limit of the sequence 
.
Solution: FTSOC, assume the limit is indeed. Then for any , there exists , such that for all , the inequality 
holds. But it's easy to see that for 
, 
Thus if we choose 
, we cannot have our desired inequality holding which contradicts our assumption, and thus the limit is not .
is not the limit of the sequence 
.Solution: FTSOC, assume the limit is indeed
. Then for any , there exists , such that for all , the inequality 
holds. But it's easy to see that for 
, 
Thus if we choose 
, we cannot have our desired inequality holding which contradicts our assumption, and thus the limit is not .Now for the other one,
Problem wrote:
Problem: Prove that the sequence 
has no limit.
Solution: Its easy to show that the point with odd numbers concentrate about the neighbourhood of point , and the points with even numers, about the neighbourhood of point . Hence, any neighbourhood of the point , as well as that of the point , contains infinite set of points . Let be an arbitrary real number. We can always choose a small that the 
neighbourhood of the point will not contain a certain neighbourhood of either the point or point . Then an infinite set of numbers will be found outside this neighbourhood, and that is why one cannot assert that all the numbers , beginning with a certain one, will enter the 
neighbourhood of the number . THis means, by definition, that the number is not the limit of the given sequenc. But is an arbitrary number, hence no number is the limit of this sequence.
has no limit.Solution: Its easy to show that the point
with odd numbers concentrate about the neighbourhood of point , and the points with even numers, about the neighbourhood of point . Hence, any neighbourhood of the point , as well as that of the point , contains infinite set of points . Let be an arbitrary real number. We can always choose a small that the 
neighbourhood of the point will not contain a certain neighbourhood of either the point or point . Then an infinite set of numbers will be found outside this neighbourhood, and that is why one cannot assert that all the numbers , beginning with a certain one, will enter the 
neighbourhood of the number . THis means, by definition, that the number is not the limit of the given sequenc. But is an arbitrary number, hence no number is the limit of this sequence.To finish with this part, let's have a teaser problem (solution left to the reader to prove
)
Problem wrote:
Prove that the sequence ![$x_n = \sqrt[n]{a}$](//latex.artofproblemsolving.com/1/5/5/155c7f1882d5dd0146a93960b40128cfe2324496.png)
, where 
, converges to , i.e., ![$$\lim_{n \to \infty} \sqrt[n]{a} = 1.$$](//latex.artofproblemsolving.com/d/0/e/d0e9728df4ee6da4c8d247f60f59d7b61b1e7c2c.png)
![$x_n = \sqrt[n]{a}$](http://latex.artofproblemsolving.com/1/5/5/155c7f1882d5dd0146a93960b40128cfe2324496.png)
, where 
, converges to , i.e., ![$$\lim_{n \to \infty} \sqrt[n]{a} = 1.$$](http://latex.artofproblemsolving.com/d/0/e/d0e9728df4ee6da4c8d247f60f59d7b61b1e7c2c.png)
This post has been edited 2 times. Last edited by kamatadu, Aug 7, 2022, 6:49 PM
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March 2025
February 2025