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Thread: Limit for series

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

    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

    $\displaystyle =\frac{n!}{n!} \times \frac{1}{2+\frac{1}{n!}}=\frac{1}{2+\frac{1}{n!}}$
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  3. #3
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    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 $\displaystyle \lim_{n \rightarrow \infty} n^2$
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