limiting value

Original post by SweetCherry1

in my book the it says to work out the limiting of n^2 + 4n divided by n +8 as n tends towards infinity . THe answer is one. I just don't understand why you get rid of the 4n??

What do you mean? n isn't a common factor, so you can't get rid of anything :confused:

What have you tried so far?

Reply 2

OP

Original post by Arithmeticae

What do you mean? n isn't a common factor, so you can't get rid of anything :confused:

What have you tried so far?

well they got rid of 4n and 8 because the answer is n^2 over n^2 which is one

Reply 3

Original post by SweetCherry1

well they got rid of 4n and 8 because the answer is n^2 over n^2 which is one

Do you mean

n^2 + \dfrac{4n}{n+8}

or

\dfrac{n^2+4n}{n+8}

?

Edit - The answer for both is infinity anyway :s-smilie:

(edited 9 years ago)

Reply 4

OP

Original post by Arithmeticae

Do you mean

n^2 + \dfrac{4n}{n+8}

or

\dfrac{n^2+4n}{n+8}

?

Edit - The answer for both is infinity anyway :s-smilie:

the second one and the answer is one- my teacher told me to get rid of the 8 because its irrelevant when n tends towards infinity. but I don't understand why you get rid of the 4n :confused:

how can the limiting value be infinity??

(edited 9 years ago)

Reply 5

Wolfram says infinity as well...

Original post by SweetCherry1

the second one and the answer is one- my teacher told me to get rid of the 8 because its irrelevant when n tends towards infinity. but I don't understand why you get rid of the 4n :confused:

Reply 6

Krollo

17

Original post by Arithmeticae

Wolfram says infinity as well...

As does L'Hopital's rule, which is always useful in these types of questions (though apparently you can't use it for the main method, though it's good for a quick check).

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Reply 7

OP

Wolfram says infinity as well...

Original post by Arithmeticae

the limiting value cant be infinity- what are you saying infinity is- n?

Reply 8

Original post by Krollo

As does L'Hopital's rule, which is always useful in these types of questions (though apparently you can't use it for the main method, though it's good for a quick check).

Posted from TSR Mobile

That's how I checked it as well, it seems a bit over the top for A level though :/

Original post by SweetCherry1

the limiting value cant be infinity- what are you saying infinity is- n?

As n tends to infinity, the function also tends to infinity.

Reply 9

davros

16

Original post by SweetCherry1

in my book the it says to work out the limiting value of n^2 + 4n divided by n +8 as n tends towards infinity . THe answer is one. I just don't understand why you get rid of the 4n??

The answer isn't one, the answer is infinity.

Are you sure you've copied the question down properly - is it supposed to be n^2 + 8 or something on the denominator?

Reply 10

OP

Original post by Arithmeticae

That's how I checked it as well, it seems a bit over the top for A level though :/

As n tends to infinity, the function also tends to infinity.

this isn't A-level its GCSE. THe answer is one- for all the questions I have beend doing on limiting values the answer is never infinity. The question is what is the limiting value- not what it tends towards

(edited 9 years ago)

Reply 11

rs232

Rearrange to get this

\Large y= x-4+\frac{32}{x+8}

and as x tends to infinity

\Large y \rightarrow x-4

Reply 12

OP

I don't think u guys understand my question. Im only doing GCSE it shouldn't be that complicated

Reply 13

Original post by SweetCherry1

this isn't A-level its GCSE. THe answer is one- for all the questions I have beend doing on limiting values the answer is never infinity.

Unless you've written it down wrong, the limit is definitely not 1... :s-smilie:

The question is what is the limiting value- not what it tends towards

AFAIK they're the same thing, aren't they... :confused:

Reply 14

OP