# Thread: derivation of a limit problem .. algebra help needed!

1. ## derivation of a limit problem .. algebra help needed!

Hello all,

I am trying to understand whats going on algebra wise here,

(6n^4+2n^3) / (7n-11^4)

from here i needed to eliminate the n^4 terms, so;

((6n^4+2n^3) / (7n-11^4)) * (1/n^4)/(1/n^4) ..

so far so good, but apparantly this all then simplifies to :

(6+(2/n))/((7/n^3)-11)

I cannot work out how we arrive at this answer.

Help would be greatly appreciated,

kind thanks

2. ## Re: derivation of a limit problem .. algebra help needed!

You are finding the limit as n → what?

3. ## Re: derivation of a limit problem .. algebra help needed!

i know that 2/n and 7/n^3 tend to zero; given by the reciprocal rule, thus this problem converges to 6/-11 . i get the final answer, its the algebra manipulation beforehand thats bothering me.

thanks

4. ## Re: derivation of a limit problem .. algebra help needed!

Originally Posted by euphmorning
i know that 2/n and 7/n^3 tend to zero; given by the reciprocal rule, thus this problem converges to 6/-11 . i get the final answer, its the algebra manipulation beforehand thats bothering me.
Look! The point is the expression (6n^4+2n^3) / (7n-11^4) is very hard to read.
The limit as $n\to \infty$ of that is $\infty$.
So I think you have posted something you don't want.

5. ## Re: derivation of a limit problem .. algebra help needed!

my problem was concerned with understanding the algebra manipulation. sorry if my problem has inadvertantly confused .

6. ## Re: derivation of a limit problem .. algebra help needed!

is this the expression you're dealing with?

$\frac{6n^4+2n^3}{7n-11n^4}$

are you trying to find the limit as $n \to \infty$ ?

if so, take care in proofing your post.

7. ## Re: derivation of a limit problem .. algebra help needed!

Originally Posted by skeeter
is this the expression you're dealing with?

$\frac{6n^4+2n^3}{7n-11n^4}$

are you trying to find the limit as $n \to \infty$ ?

if so, take care in proofing your post.
That is clearly not OP.

It was $\frac{6n^4+2n^3}{7n-11^4}$

That is why I asked for a clarification!

8. ## Re: derivation of a limit problem .. algebra help needed!

Originally Posted by skeeter
is this the expression you're dealing with?

$\frac{6n^4+2n^3}{7n-11n^4}$

are you trying to find the limit as $n \to \infty$ ?

if so, take care in proofing your post.

thanks, yes thats the expression i'm dealing with; as n goes to infinity.

9. ## Re: derivation of a limit problem .. algebra help needed!

Yes, so one thing you could do is divide the numerator and the demoninator by n^4.

This would leave you with (6 + 2/n)/(7/n^3 - 11)

now as n approaches infinity, the numerator approaches 6. This is because, as n approaches infinity, 2/n goes to 0.
and also, as n approaches infinity, the denominator approaches -11. This is because, as n approaches infinity 7/n^3 also approaches 0.

and hence you are left with 6 in the numerator and -11 in the demoninator, and thus the limit as n approaches infinity is -6/11

oh that wasn't the original post, but if that's what was intended then there you go

10. ## Re: derivation of a limit problem .. algebra help needed!

\displaystyle \begin{align*} \frac{6n^4 + 2n^3}{7n - 11n^4} &= \left(\frac{6n^4 + 2n^3}{7n - 11n^4}\right) \left(\frac{\frac{1}{n^4}}{\frac{1}{n^4}}\right) \\ &= \frac{\frac{1}{n^4}(6n^4 + 2n^3)}{\frac{1}{n^4}(7n - 11n^4)} \\ &= \frac{\frac{6n^4}{n^4} + \frac{2n^3}{n^4}}{\frac{7n}{n^4} - \frac{11n^4}{n^4}} \\ &= \frac{6 + \frac{2}{n}}{\frac{7}{n^3} - 11} \end{align*}

Now make $\displaystyle n \to \infty$.

11. ## Re: derivation of a limit problem .. algebra help needed!

thanks, that neat break down really has helped. i'm just stuck with how you reach the last part. i think this is just some basic algebra i'm not quite getting! ie. where does the 2/n and 7/n^3 expressions come from?

12. ## Re: derivation of a limit problem .. algebra help needed!

Originally Posted by euphmorning
thanks, that neat break down really has helped. i'm just stuck with how you reach the last part. i think this is just some basic algebra i'm not quite getting! ie. where does the 2/n and 7/n^3 expressions come from?
What is $\displaystyle \frac{n^3}{n^4}$ equal to? $\displaystyle \frac{n^3}{n^4} = \frac{n\cdot n\cdot n}{n\cdot n \cdot n \cdot n} = \frac{1}{n}$.

It's basic index laws, $\displaystyle \frac{n^a}{n^b} = n^{a-b}$

13. ## Re: derivation of a limit problem .. algebra help needed!

Thats making perfect sense now. many thanks!