How to solve limit of $(n-1)\left[1-\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}\right]$? I'm trying to find the solution to the following limit:
$$
\lim_{n\rightarrow\infty}(n-1)\left[1-\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}\right]\ \mathrm{.}
$$
I've tried decomposing the limit and writing it as
$$
\lim_{n\rightarrow\infty}(n-1)\left[1-\left(1-\frac{\lambda}{n}\right)^{n}\left(1+\frac{\lambda}{n}\right)^n\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{-1}\right]\ \mathrm{,}
$$
so I can use that
$$
\lim_{n\rightarrow\infty}\left(1\pm\frac{\lambda}{n}\right)^{n}=e^{\pm\lambda}\mathrm{,}
$$
but the problem seems to be that I cannot split the products or the sums within the limit into separate limits as $\lim_{n\rightarrow\infty}(n-1)$ does not converge. The answer should be $\lambda^2$ (this is also what Wolfram gives me), but I don't see a way to get to that answer myself. I'd rather not resort to the Laurent series, as this is also a nightmare to compute. Can anyone help?
 A: We have that by $\log (1+x)=x+O(x^2)$ and $e^x=1+x+O(x^2)$
$$\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}=e^{(n-1)\log\left(1-\left(\frac{\lambda}{n}\right)^2\right)}=e^{-\frac{(n-1)\lambda^2}{n^2}+O\left(\frac1{n^3}\right)}=1-\frac{(n-1)\lambda^2}{n^2}+O\left(\frac1{n^3}\right)$$
therefore
$$(n-1)\left[1-\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}\right]=\frac{(n-1)^2\lambda^2}{n^2}+O\left(\frac1{n^2}\right) \to \lambda^2$$
A: Note that\begin{multline}\lim_{n\to\infty}(n-1)\left(1-\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}\right)\\=\lim_{n\to\infty}n\left(1-\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}\right)-\lim_{n\to\infty}\left(1-\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}\right)\end{multline}(as long as both limits exist), but$$\lim_{n\to\infty}\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}=\lim_{n\to\infty}\left(1-\left(\frac\lambda n\right)^2\right)^n\left(1-\left(\frac\lambda n\right)^2\right)^{-1}=1,$$and therefore$$\lim_{n\to\infty}\left(1-\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}\right)=0$$So, all you need is to compute$$\lim_{n\to\infty}n\left(1-\left(1-\left(\frac\lambda n\right)^2\right)^{n-1}\right),$$which is equal to$$\lim_{h\to0}\frac{1-\left(1-\lambda^2h^2\right)^{1/h}}h\tag1$$(again, if it exists). Now, you can use the fact that$$\left(1-\lambda^2h^2\right)^{1/h}=e^{\log(1-\lambda^2h^2)/h}$$and that, near $0$, you have$$\frac{\log(1-\lambda^2h^2)}h=-\lambda^2h-\frac{\lambda ^4 h^3}{2}-\frac{\lambda^6 h^5}{3}+\cdots$$and therefore$$e^{\log(1-\lambda^2h^2)/h}=1-\lambda ^2 h+\frac{\lambda ^4h^2}{2}+\left(-\frac{\lambda^6}6-\frac{\lambda^4}2\right) h^3+\cdots$$It follows from this that $(1)$ is equal to $\lambda^2$, and then so is the limit of your sequence.
