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

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

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- Oct 22nd 2008, 10:52 AMRetromingentLimit 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 AMMoo
- Oct 22nd 2008, 11:04 AMterr13
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 $\displaystyle \lim_{n \rightarrow \infty} n^2$