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

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In summary, the concept of infinity can be difficult to understand, especially when trying to apply finite thinking to it. It is important to keep in mind that infinity is a concept that is still being explored and understood by mathematicians and scientists. When dealing with limits involving infinity, it is essential to fix one variable and let the other approach infinity, rather than changing both variables simultaneously.
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student34
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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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  • #2
##\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.
 
  • #3
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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  • #4
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.
 
  • #5
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.
 

1. What is the limit of x^n/(n) as n approaches infinity?

The limit of x^n/(n) as n approaches infinity is equal to 0. This is because as n gets larger and larger, the denominator (n) becomes significantly larger than the numerator (x^n), making the fraction approach 0.

2. Can the limit of x^n/(n) be solved using L'Hopital's rule?

Yes, the limit of x^n/(n) can be solved using L'Hopital's rule. This rule states that if the limit of a fraction results in an indeterminate form (such as 0/0 or infinity/infinity), then taking the derivative of the numerator and denominator separately and evaluating the limit again can solve the problem.

3. Is the limit of x^n/(n) always equal to 0?

No, the limit of x^n/(n) is not always equal to 0. It depends on the value of x and the exponent n. For example, if x = 2 and n = -2, then the limit would be equal to 4. However, as n approaches infinity, the limit will always approach 0.

4. Can the limit of x^n/(n) be solved without using L'Hopital's rule?

Yes, the limit of x^n/(n) can be solved without using L'Hopital's rule. One method is to use the fact that x^n = (x^1)^n = x^(n*1) and then apply the limit law for powers, which states that the limit of a power is equal to the power of the limit. In this case, the limit would be equal to x^0, which is equal to 1. Another method is to use the Squeeze Theorem, which states that if two functions have the same limit as x approaches a, and a third function is always between these two functions, then the third function also has the same limit as x approaches a.

5. What is the relationship between the exponent n and the limit of x^n/(n)?

The relationship between the exponent n and the limit of x^n/(n) is that as n approaches infinity, the limit will always approach 0. This is because as n gets larger, the denominator (n) becomes significantly larger than the numerator (x^n), causing the fraction to decrease and approach 0. However, for specific values of n, such as n = 1 or n = -1, the limit will be equal to 1/x, which is not necessarily 0. Additionally, for negative values of n, the limit may not exist because the fraction would result in a negative exponent, which is undefined.

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