(adsbygoogle = window.adsbygoogle || []).push({}); Application on the limit definition of "e"

Hi, I have known that:

(i)[itex] (1+\frac{a}{n})^n=((1+\frac{a}{n})^\frac{n}{a})^a\to e^a[/itex]

(ii)[itex] (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[/itex]

With above two facts, I wanted to show [itex](\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}[/itex] as n goes to infinity, for a fixed positive real t.

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

[itex](\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}[/itex]

[itex]=(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[/itex] 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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# Application on the limit definition of e

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