Application on the limit definition of e

fblues
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Application on the limit definition of "e"

Hi, I have known that:
(i)(1+\frac{a}{n})^n=((1+\frac{a}{n})^\frac{n}{a})^a\to e^a
(ii)(1-\frac{1}{n})^n=(\frac{n-1}{n})^n=(\frac{1}{\frac{n}{n-1}})^{(n-1)+1}=(\frac{1}{1+\frac{1}{n-1}})^{(n-1)}\cdot (\frac{1}{1+\frac{1}{n-1}}) \to \frac{1}{e}\cdot 1

With above two facts, I wanted to show (\frac{1}{1-\frac{t}{\sqrt{\frac{n}{2}}}})^\frac{n}{2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^\frac{t^2}{2} as n goes to infinity, for a fixed positive real t.

However, I am continuously getting e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2} instead of above result and could not find the reason on the following my argument:

(\frac{\sqrt{\frac{n}{2}}}{\sqrt{\frac{n}{2}}-t})^\frac{n}{2}=(\frac{(\frac{\sqrt{\frac{n}{2}}}{t}-1)+1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^\frac{n}{2}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t+\sqrt{\frac{n}{2}}t}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2+t^2}
=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{t^2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2}\cdot 1 as n goes to infinity.

It would be very appreciative if you let me know my mistake.
Thank you very much.
 
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Hi fblues. How can there be an ##n## on the right hand side, after you have taken the limit of ##n \to \infty##?
 
I think the statement that you want is something like

\lim_{n\to \infty} \left(\frac{1}{1-t/\sqrt{n/2}} \right)^{n/2} e^{-\sqrt{n/2} t} = e^{t^2/2}

(I don't know if this is the correct statement, but is what your statement should look similar to).
 
To. CompuChip:
Thank you for letting me know. I tried to split the part that I don't know from the original problem and made a mistake during this procedure. BTW, it seems Office_Shredder knows the original one.

To. Office_Shredder:
Yes. The problem is from "the mgf of Chi_sq(n) becomes the mgf of normal(0,1) as n goes to infinity." I think the general approach is use of Taylor expansion. But, I tried to employ the limit definition of e. Do you have an idea for this?
 
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