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Limits, tending to infinity etc. (A-level?)

Cannot do this! Going crazy!!

I have *no* idea what this question is asking me to do!

Find the limit, as n tend to infinity, of:

3(1 + (0.2)^n)

Thanks in advance!! :smile:

Reply 1

Speleo

7

As n gets bigger, what happens to 3(1 + (0.2)^n)?

Reply 2

El Matematico

limn->infinity (3 +3(0.2)^n)

3(1/5)^n -> 0 as n -> infinity

So thie limit is just 3.

Reply 3

Codefusion

1

since 0.2 is < 1 , multiplying it by it self makes it smaller.
what is the number that this expression comes close to(as you increase n) ,
but could never reach.

Reply 4

*Katie*

OP

1

Thanks

But then this one needs a completely different method, right? For this one:
5n^2 + 4n + 3 / (2n^2 + 1)

I wrote it in the form 1 + (3n+1)(n+1)/(2n^2+1)
but that doesn't help. I'm guessing it needs to be in the log form e^n? Except I can't cancel down the last bit...any help would be *greatly* appreciated!!

Reply 5

davros

16

Try writing 5n^2 + 4n + 3 as a multiple of the denominator plus some other term (which will be of the form an+b). Then the multiple stays constant as n varies and the remaining bit will be a fraction with an+b in the numerator and 2n^2+1 in the denominator, and this will tend to 0 as n goes to infinity.

So, your limit is just the multiple you worked out originally.

Reply 6

*Katie*

OP

1

davros

Try writing 5n^2 + 4n + 3 as a multiple of the denominator plus some other term (which will be of the form an+b). Then the multiple stays constant as n varies and the remaining bit will be a fraction with an+b in the numerator and 2n^2+1 in the denominator, and this will tend to 0 as n goes to infinity.

So, your limit is just the multiple you worked out originally.

Why would it be in the form (an+b)? Surely that would require the denominator to be linear (i.e. 2n+1, not (2n^2+1). Because it is a quadratic, this is what happens:

(5/2n + 3/4) remainder 2 1/4. :confused:

And I factorised it after taking 2n^2 from the beginning into (3n+1)(n+1), but I couldn't factorise the bottom, so it didn't cancel! :don'tknow:
I'm so lost, PLEASE if you have an idea, post!

Reply 7

steve10

7

*Katie*

Thanks

But then this one needs a completely different method, right? For this one:
5n^2 + 4n + 3 / (2n^2 + 1)

I wrote it in the form 1 + (3n+1)(n+1)/(2n^2+1)
but that doesn't help. I'm guessing it needs to be in the log form e^n? Except I can't cancel down the last bit...any help would be *greatly* appreciated!!

The method to use here is to divide top and bottom by the greatest power in either.
The greatest power is n². So divide top and bottom by n². giving

(5 + 4/n + 3/n²) / (2 + 1/n²)

as n get big, then 1/n and 1/n² get small. In the limit they tend to zero. So your limit is (5 + 0 + 0) / (2 + 0) = 5/2

Reply 8

*Katie*

OP

1

steve10

The method to use here is to divide top and bottom by the greatest power in either.
The greatest power is n². So divide top and bottom by n². giving

(5 + 4/n + 3/n²) / (2 + 1/n²)

as n get big, then 1/n and 1/n² get small. In the limit they tend to zero. So your limit is (5 + 0 + 0) / (2 + 0) = 5/2

I really can't thank you enough! It's so frustrating not being able to do maths, especially when you need it at uni. I have no idea what I'm going to do :eek:

Reply 9

davros

16

Are you at Uni already and struggling with some maths, or trying to do some A-level type maths in preparation for a Uni course??

Reply 10

giwtis10

Well, you should learn you theory! (look who is talking now!)
if n tend to infinity and aE(belongs)R then a^n=0

Reply 11

davros

16

What I was getting at with my earlier post, just to clarify, was that we can write:

5n^2 + 4n +3 = 2.5(2n^2 + 1) + 0.5 + 4n

so (5n^2 + 4n + 3)/(2n^2 + 1) = 2.5 + ((0.5 + 4n)/(2n^2 + 1))

The 2.5 stays constant whatever happens to n, and the 2nd term gets closer and closer to 0 as n increases because the numerator just has an n in it and the denominator has an n^2.

Therefore the limit is 2.5, which is the same answer Steve reached with a slightly different method.

Limits, tending to infinity etc. (A-level?)

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