Limit for series

**Retromingent** · Oct 22nd 2008, 10:52 AM

$\lim_{x\to\infty}\frac{n!}{2n! + 1} = \frac{1}{2}$

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

**Moo** · Oct 22nd 2008, 10:57 AM

Originally Posted by Retromingent

$\lim_{x\to\infty}\frac{n!}{2n! + 1} = \frac{1}{2}$

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

Factor out $n!$ from the numerator and the denominator

$=\frac{n!}{n!} \times \frac{1}{2+\frac{1}{n!}}=\frac{1}{2+\frac{1}{n!}}$

**terr13** · Oct 22nd 2008, 11:04 AM

I was going to say that since $n!$ will obviously be much larger than the constant, you only need to look at the coefficients of $n!$ . The same holds true for powers, you only need to look at the highest power, since $\lim_{n \rightarrow \infty} n^3$ rises much faster than $\lim_{n \rightarrow \infty} n^2$

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