# $\lim_{n \to \infty} \left| \cos \left( \frac{\pi}{4(n-1)} \right) \right|^{2n-1}$

I need Help evaluating the limit of $$\lim_{n \to \infty} \left| \cos \left( \frac{\pi}{4(n-1)} \right) \right|^{2n-1} = L$$

I already know that $L = 1$, but I need help showing it.

The best idea I could come up with was to take the series representation of cosine. $$\lim_{j,n \to \infty} \left| 1 - \left( \frac{\pi}{4(n-1)} \right)^2 \frac{1}{2!} + ....+ \frac{(-1)^j}{(2j)!}\left( \frac{\pi}{4(n-1)} \right)^{2j} \right|^{2n-1} = L$$

All lower order terms go to zero leaving:

$$\lim_{j,n \to \infty} \left|\frac{1}{(2j)!}\left( \frac{\pi}{4(n-1)} \right)^{2j} \right|^{2n-1} = L$$

But this doesn't really seem like I am any closer. How do I proceed? Obviously L'Hospitals rule will occur eventually. Hints?

Put $$L = \lim_{n \to \infty} \left| \cos \left( \frac{\pi}{4(n-1)} \right) \right|^{2n-1}$$ Then $$\log(L) = \lim_{n \to \infty} (2n-1)\log\left( \cos \left( \frac{\pi}{4(n-1)} \right) \right).$$ This is an $0\cdot \infty$ indeterminate form. Put the $2n -1$ in the denominator and invoke L'hospital.
Your idea of using, for a sufficiently large value of $n$, different Taylor expansions looks good to me. Built around $x=0$, we have $\cos(x)=1-\frac{x^2}{2}+O\left(x^3\right)$. On the other hand, $(1-y)^k$ is almost $(1-k y)$. So, replace $y$ by $x^2$, $k$ by $(2n-1)$,$x$ by $\frac{\pi }{4 (n-1)}$ and you end with $$1-\frac{\pi ^2 (2 n-1)}{32 (n-1)^2}$$ which, for a sufficiently large value of $n$ is almost $$1-\frac{\pi ^2}{16 n}$$Now, push $n$ to infinity.