limit of (10^n)/(n!) as n->infinity
The answer is zero. Can you prove it?
What you need to do is prove that for every epsilon > 0, you can find an N such that for all n > N you have that
10^n/n! < epsilon
You can write
10^n/n! = Product from k=1 to n of 10/k
In this product the first ten factors are larger than 1. For k > 10 the factors are smaler than 1. If we call the product of the first ten factors F, then for n > 10, we clearly have:
10^n/n! < F*10/n
because if we omit all te factors between k = 11 and k = n-1, we are clearly overestimating the product.
This means that if you take N such that
F*10/N < epsilon -------->
N > 10 F/epsilon
then we can be sure that for all n > N the product is less than epsilon.