Convergence of the following sequence: $\lim_{n\to \infty} e^{-t\sqrt{n}}(1-\frac{t}{\sqrt{n}})^{-n}=e^{\frac{1}{2}t^2}$

Question

It could be exhaustion from the amount of work that I've done today, but I'd like to prove for myself that $\lim_{n\to \infty} e^{-t\sqrt{n}}(1-\frac{t}{\sqrt{n}})^{-n}=e^{\frac{1}{2}t^2}$$

Here's what I've attempted:

Take the log of our sequence. Then we have $\lim_{n\to \infty} n\cdot t\sqrt{n}\cdot \ln(1-\frac{t}{\sqrt{n}}) \implies \lim_{n\to \infty} \ln((1-\frac{t}{\sqrt{n}})^{\sqrt{n}}) \implies \lim_{n\to \infty} \frac{\ln((1-\frac{t}{\sqrt{n}})^{\sqrt{n}})}{n^{-1}}$

To be frank, I'm not sure how to proceed from this step. I've applied L'Hopital's rule to this final step and it creates an utter mess. Every route I take seems to end in divergence, but I know this converges.

Answer 1

We have, taking log and using L'Hopital, $$\lim_{n\rightarrow\infty}-t\sqrt{n}-n\log\left(1-\frac{t}{\sqrt{n}}\right)=\lim_{n\rightarrow\infty}\frac{\left(-\frac{t}{\sqrt{n}}-\log\left(1-\frac{t}{\sqrt{n}}\right)\right)}{1/n}=\frac{1}{2}t^{2}\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\sqrt{n}-t} $$ and so your limit.