How do I calculate the limit of $\lim_{n\to \infty} (1-\frac{\theta^2}{2n^2})^{2(n+1)}$

Question

I'm going through some physics problems about polarizers and one problem is about the case where $n+1$ polarizers are stacked up and I have to look at the case where $n \to \infty$.

Now I came up for a solution for the intensity in the case of $n+1$ polarizers: $$I_{n+1}=I_0*\left(\cos^2\left(\frac{\theta}{n}\right)\right)^{n+1}$$

Doing the Taylor expansion of $\cos(x)$ I get: $$I_{n+1}=I_0*\left(1-\frac{\theta^2}{2n^2}\right)^{2(n+1)}$$

I know that the limit of this as $n \to\infty$ should be just $I_0$, but I don't really know how to get to that result. I also saw someone saying that $$\left(1-\frac{\theta^2}{2n^2}\right)^{2(n+1)} \approx e^{-\theta^2/(n+1)}$$ which would indeed give me $1$ as $n \to \infty $ but I don't want to use something that I don't fully understand how to get to.

Would be really great if someone could help me out! Thank you!

Answer 1

Take logarithms:$$\lim_{n\to\infty}2(n+1)\ln(1-\theta^2/2n^2)=\lim_{n\to\infty}2(n+1)(-\theta^2/2n^2)=\lim_{n\to\infty}(-\theta^2/n)=0,$$with repeated uses of $f(n)\sim1\to\lim_{n\to\infty}g(n)=\lim_{n\to\infty}f(n)g(n)$. So the original limi is $e^0=1$.

Answer 2

$$\lim\left(1-\frac{\theta^2}{2n^2}\right)^{2(n+1)}=\lim\left(1-\frac{\theta^2}{2n^2}\right)^{2n^2/n}\left(1-\frac{\theta^2}{2n^2}\right)^2=\left(e^{-\theta^2}\right)^{\lim 1/n}1^2=1.$$

Answer 3

Yet another way: $$ \left(1 - \frac{\theta^2}{2n^2}\right)^{2(n + 1)} = \left(1 - \frac{\theta^2}{2n^2}\right)^2 \left(1 + \frac{\theta\sqrt2}{2n}\right)^{2n} \left(1 - \frac{\theta\sqrt2}{2n}\right)^{2n} \to 1^2 \cdot e^{\theta\sqrt2} \cdot e^{-\theta\sqrt2} = 1. $$