Question

Find the limit please!

Answer 1

got anymore

Answer 2

@dimensionx ?

Answer 3

Hint : you can pull the denominator out of sum as it doesn't have the index variable s

Answer 4

lol

Answer 5

@amysparkly12 the answer should be an exact number

Answer 6

whats the choices then?

Answer 7

It's free response :/

Answer 8

okay then 907

Answer 9

\[\begin{align}\lim\limits_{N\to\infty}\sum\limits_{s=1}^N\dfrac{3s^2-5s-3}{N^3} &=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\sum\limits_{s=1}^N(3s^2-5s-3)\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\sum\limits_{s=1}^Ns^2-5\sum\limits_{s=1}^Ns-3\sum\limits_{s=1}^N1\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\dfrac{N(N+1)(2N+1)}{6}-5\dfrac{N(N+1)}{2}-3N\right)\\~\\

\end{align}\]

Answer 10

see if you can take it home

Answer 11

thanks, Is the last term 3N? it's cut off for me

Answer 12

Oh see if below shows up fully

\[\begin{align}&\lim\limits_{N\to\infty}\sum\limits_{s=1}^N\dfrac{3s^2-5s-3}{N^3}\\~\\ &=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\sum\limits_{s=1}^N(3s^2-5s-3)\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\sum\limits_{s=1}^Ns^2-5\sum\limits_{s=1}^Ns-3\sum\limits_{s=1}^N1\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\dfrac{N(N+1)(2N+1)}{6}-5\dfrac{N(N+1)}{2}-3N\right)\\~\\

\end{align}\]

Answer 13

yeah that's better.

So just wondering, the first sum is 1?

Answer 14

\[\lim_{N \rightarrow infinity}\frac{ 1 }{ N^{3} } (-5\frac{ N(N+1) }{ 2 })\] is this 0?

Answer 15

careful

Answer 16

I don't know what do do with the last 2 terms, the denominator seems like it's bigger?

Answer 17

rest is just algebra

Answer 18

\[\begin{align}&\lim\limits_{N\to\infty}\sum\limits_{s=1}^N\dfrac{3s^2-5s-3}{N^3}\\~\\ &=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\sum\limits_{s=1}^N(3s^2-5s-3)\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\sum\limits_{s=1}^Ns^2-5\sum\limits_{s=1}^Ns-3\sum\limits_{s=1}^N1\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\dfrac{N(N+1)(2N+1)}{6}-5\dfrac{N(N+1)}{2}-3N\right)\\~\\

&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\dfrac{N(N+1)(2N+1)}{2}-\dfrac{5N^2+5N}{2}-\dfrac{6N}{2}\right)\\~\\

\end{align}\]

Answer 19

combine the fractions inside parenthesis

Answer 20

the numerator will have a dergree of 3 right ?

Answer 21

Oh! So the denominators are all the same, the numerator is a degree of 3.. do I multiply the coefficients of the leading term (N^3) and divide by 2?

Answer 22

yes!

Answer 23

3000/2 = 1500?

Answer 24

how did u get 3000 ? :O

Answer 25

300/2 = 150?

Answer 26

nope

Answer 27

2/2 = 1
is that the first sum?

Answer 28

\[\begin{align}&\lim\limits_{N\to\infty}\sum\limits_{s=1}^N\dfrac{3s^2-5s-3}{N^3}\\~\\ &=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\sum\limits_{s=1}^N(3s^2-5s-3)\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\sum\limits_{s=1}^Ns^2-5\sum\limits_{s=1}^Ns-3\sum\limits_{s=1}^N1\right)\\~\\
&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(3\dfrac{N(N+1)(2N+1)}{6}-5\dfrac{N(N+1)}{2}-3N\right)\\~\\

&=\lim\limits_{N\to\infty}\frac{1}{N^3}\left(\dfrac{N(N+1)(2N+1)}{2}-\dfrac{5N^2+5N}{2}-\dfrac{6N}{2}\right)\\~\\

&=\lim\limits_{N\to\infty}\left(\dfrac{2N^3 + \text{some other lower degree terms}}{2N^3} \right)\\~\\

\end{align}\]

Answer 29

oh, so it's just 1

Answer 30

1 is \(\large \color{Red}{\checkmark }\)

Answer 31

Thanks a lot!

Answer 32

yw!