Question

Can you solve this question using series? (Do not involve Lhospital rule anywhere)

$\Rightarrow \lim_{x \to 18} \left(\cos\left(\ln\left(2\cos\frac{6\pi}{x}\right)\right)\right)^{\frac{1}{\left(\arctan\frac{x}{18} - \frac{\pi}{4}\right)^2}}$

Answer 1

$e^{-\frac{2\pi^2}{3}}$

Answer 2

The limit is of the form $1^\infty$, which is converted to $e^{\lim_{x \to a} g(x)(f(x)-1)}$. Let $x = 18+y$, so $y \to 0$.

1. Expand $\left(\arctan\frac{x}{18} - \frac{\pi}{4}\right)^2$ using Taylor series for $\arctan(1+u)$ around $u=0$. This gives $\frac{y^2}{1296}\left(1 - \frac{y}{18} + O(y^2)\right)$.

2. Expand $\cos\left(\ln\left(2\cos\frac{6\pi}{x}\right)\right) - 1$.

   a. Expand $\frac{6\pi}{x}$ as $\frac{\pi}{3} + \delta$, where $\delta = -\frac{\pi y}{54} + O(y^2)$. b. Expand $2\cos\left(\frac{\pi}{3}+\delta\right)$ using $\cos(\frac{\pi}{3}+\delta) = \frac{1}{2}\cos\delta - \frac{\sqrt{3}}{2}\sin\delta$ and series for $\sin\delta, \cos\delta$. This gives $1 - \sqrt{3}\delta - \frac{\delta^2}{2} + O(\delta^3)$. c. Expand $\ln(2\cos t)$ using $\ln(1+M) = M - \frac{M^2}{2} + O(M^3)$. This gives $-\sqrt{3}\delta - 2\delta^2 + O(\delta^3)$. d. Substitute $\delta$ in terms of $y$ into $\ln(2\cos t)$. This gives $\frac{\sqrt{3}\pi y}{54} + O(y^2)$. e. Expand $\cos(Z)-1$ using $\cos Z - 1 = -\frac{Z^2}{2} + O(Z^4)$, where $Z = \ln(2\cos t)$. This gives $-\frac{\pi^2 y^2}{1944} + O(y^3)$.

3. Multiply the series expansions of $g(x)$ and $(f(x)-1)$ and take the limit as $y \to 0$.

The limit of the exponent simplifies to $-\frac{2\pi^2}{3}$. The final answer is $e^{-\frac{2\pi^2}{3}}$.