limit problem,

[shelly89]

\(\displaystyle a_{n} = \frac{2n^{2}+n}{n^{2}+2n} \)

then the limit as n approaches infinity is ?


not sure what to do... i tired to factorise and \(\displaystyle \frac{2n+1}{1+2n} \)

but this gives me 1, the correct answer is 2?

 

[Romsek]

\(\displaystyle a_{n} = \frac{2n^{2}+n}{n^{2}+2n} \)

then the limit as n approaches infinity is ?


not sure what to do... i tired to factorise and \(\displaystyle \frac{2n+1}{1+2n} \)

but this gives me 1, the correct answer is 2?

try dividing top and bottom by n² and then let n -> infinity

 

[stapel]

\(\displaystyle a_{n} = \frac{2n^{2}+n}{n^{2}+2n} \)

then the limit as n approaches infinity is ?

Try using what you learned back in algebra for finding horizontal asymptotes.

 

[JeffM]

\(\displaystyle a_{n} = \frac{2n^{2}+n}{n^{2}+2n} \)

then the limit as n approaches infinity is ?


not sure what to do... i tired to factorise and \(\displaystyle \frac{2n+1}{1+2n} \) Well, to START WITH, you did not factor correctly.

\(\displaystyle \dfrac{2n^2 + n}{n^2 + 2n} = \dfrac{n(2n + 1)}{n(n + 2)} = \dfrac{2n + 1}{n + 2}.\)

but this gives me 1, the correct answer is 2?

A third way to solve this is by algebraic division

\(\displaystyle \dfrac{2n^2 + n}{n^2 + 2n} = 2 - \dfrac{3}{n + 2}.\)