Limit for series

• Oct 22nd 2008, 10:52 AM
Retromingent
Limit for series
$\displaystyle \lim_{x\to\infty}\frac{n!}{2n! + 1} = \frac{1}{2}$

How is this so exactly? Is $\displaystyle \infty$ plugged in? And I don't see algebraically how to cancel $\displaystyle n!$
• Oct 22nd 2008, 10:57 AM
Moo
Quote:

Originally Posted by Retromingent
$\displaystyle \lim_{x\to\infty}\frac{n!}{2n! + 1} = \frac{1}{2}$

How is this so exactly? Is $\displaystyle \infty$ plugged in? And I don't see algebraically how to cancel $\displaystyle n!$

Factor out $\displaystyle n!$ from the numerator and the denominator :)

$\displaystyle =\frac{n!}{n!} \times \frac{1}{2+\frac{1}{n!}}=\frac{1}{2+\frac{1}{n!}}$
• Oct 22nd 2008, 11:04 AM
terr13
I was going to say that since $\displaystyle n!$ will obviously be much larger than the constant, you only need to look at the coefficients of $\displaystyle n!$. The same holds true for powers, you only need to look at the highest power, since $\displaystyle \lim_{n \rightarrow \infty} n^3$ rises much faster than $\displaystyle \lim_{n \rightarrow \infty} n^2$