Is it correct to separate a limit, apply Stirling and then reattach it?

Question

I don't know if I can do this with my limit:

$$\lim _{n\rightarrow \infty }{\frac {n \left( 2\,n \right) ^{2\,n}{ {\rm e}^{-2\,n}}}{ \left( 2\,n \right) !} \left( {\frac { \left( 2\,n \right) !}{ \left( 2\,n \right) ^{2\,n}{{\rm e}^{-2\,n}}}}-{\frac { \left( 2\,n+2 \right) !}{ \left( 2\,n+2 \right) ^{2\,n+2}{{\rm e}^{-2 \,n-2}}}} \right) }= $$

$$=\lim _{n\rightarrow \infty }{\frac {n}{ \sqrt{4\,\pi \,n}} \left( { \frac { \left( 2\,n \right) !}{ \left( 2\,n \right) ^{2\,n}{{\rm e}^{- 2\,n}}}}-{\frac { \left( 2\,n+2 \right) !}{ \left( 2\,n+2 \right) ^{2 \,n+2}{{\rm e}^{-2\,n-2}}}} \right) }= $$

$$=\lim _{n\rightarrow \infty }{\frac {n \left( 2\,n \right) !}{ \left( 2 \,n \right) ^{2\,n}{{\rm e}^{-2\,n}} \sqrt{4\,\pi \,n}}}-\lim _{n \rightarrow \infty }{\frac {n \left( 2\,n+2 \right) !}{ \left( 2\,n+2 \right) ^{2\,n+2}{{\rm e}^{-2\,n-2}} \sqrt{4\,\pi \,n}}}= $$

$$=\lim _{n\rightarrow \infty }{\frac {n \sqrt{4\,\pi \,n}-n \sqrt{2\, \pi \, \left( 2\,n+2 \right) }}{ \sqrt{4\,\pi \,n}}}= \lim _{n\rightarrow \infty }\frac {\sqrt{2\,{n}^{2}}- \sqrt{2\,{n}^{ 2}+2\,n}}{ \sqrt{2}}. $$

Is "legal" to separate the limit, apply Stirling and then reattach it?

Thanks.

Answer 1

• You can separate the limit into two terms or factors if both of them converge.

• You don't need to separate them to apply Stirling's formula.

• You don't even need to apply Stirling's approximation at all, since the factorials in the numerators and denominator cancel each other almost completely out.