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Show that? #(P(N', n))/(P(N,n)) ~~ ((N')/N)^n# #(N', N, n) in ZZ^+# #(N', N )">>" n# #P(x,y) = (x!)/((x-y)!)# Stirling Approx allowed

Answer 1

Stirling's approximation says that #x! approx sqrt(2pi n) (n/e)^n = sqrt(2pi) * sqrt(n) * n^n * e^-n#

In this way, we can approximate #P(x, y)#:

#(x!)/((x-y)!) approx sqrt(2pi)/sqrt(2pi) cdot sqrt(x/(x-y)) cdot (x^x/(x-y)^(x-y)) e^(-x)/e^(-(x-y)) #

Since the second condition says that #y# is far less than #x# for our purposes, so #sqrt(x) approx sqrt(x-y) # And since all of these we know that #x# and #x-y# are far greater than #e#, it is clear that the dominating term is #P(x, y) approx (x^x)/((x-y)^(x-y)) = (x/(x-y))^x (x-y)^y approx (x/x)^x (x-y)^y# #P(x, y) approx x^y#

Returning to the original question, #(P(N', n))/(P(N, n)) approx ((N')^n)/((N)^n) = ((N')/N)^n #

completing the derivation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Show that? #(P(N', n))/(P(N,n)) ~~ ((N')/N)^n# #(N', N, n) in ZZ^+# #(N', N )"&gt;&gt;" n# #P(x,y) = (x!)/((x-y)!)# Stirling Approx allowed

Show that? #(P(N', n))/(P(N,n)) ~~ ((N')/N)^n# #(N', N, n) in ZZ^+# #(N', N )">>" n# #P(x,y) = (x!)/((x-y)!)# Stirling Approx allowed