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

Question

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

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 $e^t$.

Answer 1

Hints:

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

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

$$(3)\;\lim_{n\to\infty}a_n=a>0\implies\lim_{n\to\infty}\sqrt[n]{a_n}=1$$

Answer 2

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.

Answer 3

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.