Thread: 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 $\displaystyle n!$

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

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

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 $\displaystyle \lim_{n \rightarrow \infty} n^2$