Question about the Limit of x^n/(n)

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The discussion centers on the limit of x^n/(n!) as n approaches infinity, which is established to equal 0 for fixed x in ℝ. A participant mistakenly varied both n and x, leading to confusion about the behavior of the sequence, particularly when comparing growth rates of x^n and n!. The clarification emphasizes that when evaluating limits, one must keep x fixed while allowing n to grow. The complexities of understanding infinity are acknowledged, highlighting that it can be counterintuitive and challenging to grasp. Ultimately, the conversation reflects on the nature of infinity and its implications in mathematical limits.
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My textbook explains how the limit of x^n/(n!) as n→∞ equals 0, x ∈ ℝ. Since a(n) = x^n/(n!) and a(n+1) = x^(n+1)/((n+1)n!) and (a(n+1))/a(n) = x/(n+1) and lim n→∞ x/(n+1) = 0, then it seems obvious.

But, I went on Wolfram Alpha and I noticed that when I made x = n and put in some large values, x^n increased much faster than n! did. Is this just one of those weird outcomes of using infinity? Or why does infinity change the regularity of the sequence so much to the point that they invert?
 
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##\lim_{n \to \infty} {x^n \over n!}## means that you fix x and see what happens when n becomes very large. You don't get to change n and x at the same time.

You didn't keep n fixed, which means you ended up calculating ##\lim_{n \to \infty} {n^n \over n!}## instead. That is a very different problem.

Here's a simpler example to illustrate what I'm talking about. You wrote: ##\lim_{n \to \infty} {x \over n+1} = 0##. But if you put in ##x = n## and let both become large, you'll get a sequence approaching 1, not 0. If you're not taking a limit with respect to a symbol, then you need to leave it fixed.
 
eigenperson said:
##\lim_{n \to \infty} {x^n \over n!}## means that you fix x and see what happens when n becomes very large. You don't get to change n and x at the same time.

You didn't keep n fixed, which means you ended up calculating ##\lim_{n \to \infty} {n^n \over n!}## instead. That is a very different problem.

Here's a simpler example to illustrate what I'm talking about. You wrote: ##\lim_{n \to \infty} {x \over n+1} = 0##. But if you put in ##x = n## and let both become large, you'll get a sequence approaching 1, not 0. If you're not taking a limit with respect to a symbol, then you need to leave it fixed.

I know that I am not suppose to fix n. I just wanted to see how x^n/n! looks as n and x increase. I did this because it is very counterintuitive to me how n! can ever be larger than x^n.

For example lim n→∞ 1000/(7n) = 0 is obvious because we can see that the ratio of 1000/7n decreases as n gets larger. This does not at all happen when I increase x and n values for x^n/(n!). In fact, the ratio increases as n and x get larger.

*I made a big mistake. I edited out "difference" and put in "ratio" instead.
 
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Never mind; this whole thing is dumb. I realize that 2x/n does the same thing as x and n get larger. I guess it's just my brain having a tough time with understanding how big infinity is or what infinity is.
 
Don't beat yourself up too much.

Infinity is not actually an easy thing to understand: it's only been roughly the last one hundred years that infinity is being explored (and it still is being explored today in many ways).

In some ways infinity is intuitive and in other ways it is not if you still try and make sense of it using finite thinking and its very easy to resort to finite thinking due to many of our experiences and senses living in the finite domain.
 
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