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

How can I find the limit above? I am bit confused because of the square in the denominator. If it was $n$ instead, then the limit would have simply been $e^t$.

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up vote 3 down vote accepted


$$(1)\; \left(1-\frac{2t}{n^2}\right)^{-n/2}=\left(\left[\left(1-\frac{2t}{n^2}\right)^{n^2}\right]^{-1/2}\right)^{1/n}$$

$$(2)\;\text{ For any function}\;\;f(n)\;\;s.t.\;\;\lim_{n\to\infty}f(n)=\infty\;,\;\;\lim_{n\to\infty}\left(1+\frac x{f(n)}\right)^{f(n)}=e^x$$


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The third property is very interesting. – JohnK Dec 18 '13 at 17:32
It follows at once from $\;\sqrt[n]b\longrightarrow 1\;\;\forall\,b>0\;$ , @JohnK – DonAntonio Dec 18 '13 at 17:33
That is brilliant, thank you. – JohnK Dec 18 '13 at 17:35

Hint. Take the logarithm of your term and since $n$ is large, use a first order Taylor expansion. Multiplied by the exponent, you will find $0$ as the limit of the logarithm, then $1$ for the limit.

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Hint: rewrite your limit as $$\lim_{n\to\infty}\left[\left(1-\frac{\sqrt {2t}}{n}\right)^{-n/2}\left(1+\frac{\sqrt {2t}}{n}\right)^{-n/2}\right]$$ and use the product rule.

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And if $\;t<0\;$ ? I doubt the OP's interested in complex limits... – DonAntonio Dec 18 '13 at 17:34
This is an interesting approach but if $t<0$ it can get messy. Thank you nevertheless. – JohnK Dec 18 '13 at 17:39

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