- #1
Kuno
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Homework Statement
Why does the limit as n -> infinity of [3^(n+1)]/(n+1)!] * n!/(3^n) equal
the limit as n -> infinity of 3/(n+1)?
Homework Equations
The Attempt at a Solution
I have never encountered this before.
yenchin said:Forget about the limit, how do you simplify:
[3^(n+1)]/(n+1)! * [n!/(3^n)]
(LOL, Curious3141 is faster than me)
The limit of 3/(n+1) as n approaches infinity is 0. This is because as n gets larger and larger, the denominator (n+1) becomes significantly larger than the numerator (3), making the fraction approach 0.
To calculate the limit of 3/(n+1), you can use the limit laws and algebraic manipulation to rewrite the expression into a form where you can easily see the limit. In this case, you can divide both the numerator and denominator by n, which will give you the expression 3/n. Then, you can apply the limit law that states the limit of a constant (in this case, 3) is equal to the constant. Therefore, the limit of 3/(n+1) is equal to 3/n, which approaches 0 as n approaches infinity.
Yes, the limit of 3/(n+1) is a finite value. As n approaches infinity, the fraction approaches 0, which is a finite number. This means that the limit of 3/(n+1) exists and is equal to 0.
No, L'Hopital's rule cannot be used to find the limit of 3/(n+1). L'Hopital's rule can only be used for limits involving indeterminate forms such as 0/0 or infinity/infinity. In this case, the limit of 3/(n+1) is not an indeterminate form, so L'Hopital's rule cannot be applied.
The factorial does not affect the limit of 3/(n+1). As n approaches infinity, the factorial term becomes insignificant compared to the n term, so it does not affect the overall limit. This is because factorial grows much faster than a polynomial function (such as n), so it becomes negligible in the limit calculation.