- #1
nhmllr
- 185
- 1
Here's the weird limit:
lim n[itex]\rightarrow[/itex] [itex]\infty[/itex] aCn/na = 1/(a!)
Don't ask how I thought this up, but let me explain my reasoning.
Let's say that a=3. Then aCn = (n)(n-1)(n-2)
And because n is approaching infinity, could it be (n)(n)(n) = n3?
I wouldn't normally think of things like this but if it's true then it helps me out.
The only thing is that (n)(n-1)(n-2)= n3 +something*n2+something*n
And when I graphed it it looked like it approaches zero.
I've never learned limits formally. Thanks.
EDIT: aCn is a Choose n, or n!/(a![n-a]!)
EDIT: Wolfram alpha is amazing http://www.wolframalpha.com/input/?i=lim+(x!/(3!(x-3)!x^3))+as+x->+infinity
Sorry for not trying this before posting. I'd still be interested in other replies as to why this is true, though
lim n[itex]\rightarrow[/itex] [itex]\infty[/itex] aCn/na = 1/(a!)
Don't ask how I thought this up, but let me explain my reasoning.
Let's say that a=3. Then aCn = (n)(n-1)(n-2)
And because n is approaching infinity, could it be (n)(n)(n) = n3?
I wouldn't normally think of things like this but if it's true then it helps me out.
The only thing is that (n)(n-1)(n-2)= n3 +something*n2+something*n
And when I graphed it it looked like it approaches zero.
I've never learned limits formally. Thanks.
EDIT: aCn is a Choose n, or n!/(a![n-a]!)
EDIT: Wolfram alpha is amazing http://www.wolframalpha.com/input/?i=lim+(x!/(3!(x-3)!x^3))+as+x->+infinity
Sorry for not trying this before posting. I'd still be interested in other replies as to why this is true, though
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