# Finding limit (basic question)

• Jan 1st 2008, 10:44 AM
tabularasa
Finding limit (basic question)
Lets say I have:
$\displaystyle \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= ?$

I know that the limit of ƒ of n, as n approaches infinity, is 2 but how I can get that right answer with the mathematical way? Of course I can quess answer with some big numbers but that's not what I'm looking for.

• Jan 1st 2008, 10:51 AM
bobak
$\displaystyle \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}$

$\displaystyle \lim_{n\to\infty}\frac{2n^2 + 1 - 2n +2n}{n^2 - n}$

$\displaystyle \lim_{n\to\infty}\frac{2(n^2-n) +1+ 2n}{n^2 - n}$

$\displaystyle \lim_{n\to\infty} \left ( 2 +\frac{1+ 2n}{n^2 - n} \right )$

$\displaystyle 2 + \lim_{n\to\infty}\frac{1+ 2n}{n^2 - n}$

$\displaystyle 2 + 0 = 2$
• Jan 1st 2008, 11:01 AM
Krizalid
Quote:

Originally Posted by tabularasa
Lets say I have:
$\displaystyle \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= ?$

Why don't you divide top & bottom by $\displaystyle n^2$?
• Jan 1st 2008, 11:03 AM
polymerase
Just divide to and bottom by the highest power, $\displaystyle n^2$

$\displaystyle \lim \frac{2+\frac{1}{n^2}}{1-\frac{1}{n}}$ then when the limit goes to infinity the fractions go to zero....

Hence, $\displaystyle \lim_{n\to\infty}\frac{2n^2 + 1}{n^2 - n}= \lim_{n\to\infty} \frac{2+\frac{1}{n^2}}{1-\frac{1}{n}}=2$