Scribe Post 05/18/07 Francisco and Miguel

(In this period, we continued talking about limits)
This period, we started as usual with a POD:

tn= 1/n find:

A.- Lim tn=
n --> oo

B.- Lim tn=
n --> -oo

C.- Lim tn=
n --> 0

How to do it:

On the firt problem, n is aproaching to infinity, so we have
to make a table of values with positive integers, for example:

n 1/2 1 2 3 4 5
tn 2 1 1/2 1/3 1/4 1/5

With this table of values, we can do the graph to check and prove the answer.

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So, with that we can know:
Lim tn= 0
In In this problem, the answer is 0, beacuse n will never pass 0.

On the next problem, the metod to solve it is very similar, the difference is that on here,
n is approaching to -oo ( negative infinity). So the table of values have to be like this:

n -1/2 -1 -2 -3 -4
tn -2 -1 -1/2 -1/3 -1

With this table, we can see a graph like this:

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So with this graph, we can conclude that:
Lim tn= 0
n --> -oo

This answer is beacuse the n never going to pass 0, so the limit is 0

The last problem was a littlebit different than the previous problems,
the difference is that in this problem, n is approaching 0, so the table of values
should be similar than this one:

n .01 .001 .0001 .00001
tn 100 1000 10000 100000

With this values, the graph should be:

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Lim tn= DNE
n --> 0

This answer, is beacuse n will never reach 0 on one side and neither the other side.

Very Important Note: When we finished the POD, mr. alcantara told us that some one in
AFRICA is watching our blog (.. very interesting...)

Later, we started doing some exercices about limits.

* Solve without using calculator:

1.- tn= cos (n) Lim tn=
n --> 00

To solve this problem, we have to know the shape of a cosine graph:

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Knowing the shape of the graph, we can see that it doesnt have limits, so we have to answer:
DNE that means: The Limit Does Not Exist

tn = Cos (n) Lim tn= DNE
n --> oo



2.- tn= Ln (n) Lim tn=
n --> oo


To solve this problem, we have to know the shape of Lnx, so we can know it easily if we know the shape of e^x, because Ln is the inverse function of e^x, so we can know the next graph:

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With this graph, we can conclude that:

tn= Ln (n) Lim tn= oo
n --> oo


Nearly of the end of the period, Mr. Alcantara puted us another problem:

* Find the limit ( accurate to 3 decimals) of the sequence formed by taking the ratio
of successive terms of the fibonacci sequence:

Lim Fn/Fn-1=
n --> 00

To answer this problems, firts we should know the Fibonacci sequence, that is:
1,1,2,3,5,8,12,21,31, ...

So we have to make a table of values, for example:
n 2 3 4 5 6
tn 1 2 3/2 5/3 8/5

With this values, we can do this graph:

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As we can see on the graph, the integers are not passing 2 and 1, but as Mr. Alcantara told us,
the numbers are going to stay between a specific number, and that number is:
1.618

So the answer of the problem is:

Lim Fn/Fn-1= phi
n--> 00


When we finished this problem, the period was OVER.....

Remember: Quiz Wednesday, about : Sequences, Series, Limits.