Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $e^t$.

share|cite|improve this question

3 Answers 3

up vote 3 down vote accepted


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

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


share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.