Finding limit (basic question)

**tabularasa** · January 1st 2008, 10:44 AM

Lets say I have:
$ \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.

Could someone please help me on this...

**bobak** · January 1st 2008, 10:51 AM

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

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

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

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

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

$2 + 0 = 2 $

**Krizalid** · January 1st 2008, 11:01 AM

Originally Posted by tabularasa

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

Why don't you divide top & bottom by $n^2$ ?

**polymerase** · January 1st 2008, 11:03 AM

Just divide to and bottom by the highest power, $n^2$

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

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

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