= Solution
For fixed $x>1$, write $a=\sqrt{x^2-1}$ and $A=x+a$. The maximum of the integrand occurs at $\theta=0$, and
$$
\log(x+a\cos\theta)=\log A-\frac{a}{2A}\theta^2+O(\theta^4).
$$
Here is the required endpoint <Laplace method> derivation. Outside a fixed neighborhood of zero, the integrand is exponentially smaller than $A^n$. Inside that neighborhood, use $\theta=(A/(na))^{1/2}u$. The quadratic term gives $e^{-u^2/2}$ and the higher terms vanish in the scaled limit. The <Gaussian integral> on the half-line is $\sqrt{\pi/2}$, so
$$
\boxed{P_n(x)\sim\frac{(x+\sqrt{x^2-1})^{n+1/2}}{\sqrt{2\pi n\sqrt{x^2-1}}},\qquad x>1\ \text{fixed}.}
$$
This is the <Hyperbolic Debye asymptotic for Legendre polynomials>. At the endpoint the integrand is identically one, giving \b[$P_n(1)=1$ exactly].
The localization width in the <Laplace method> is $O((na)^{-1/2})$. It becomes order one when $n\sqrt{x-1}=O(1)$, so the <Legendre boundary layer at x equals one> has \b[$q=2$]. With $x=1+\nu/n^2$ and fixed $\nu\ge0$,
$$
n\log(x+\sqrt{x^2-1}\cos\theta)=\sqrt{2\nu}\cos\theta+O(n^{-1})
$$
uniformly in $\theta$. The boundary-layer answer is therefore
$$
\boxed{P_n(1+\nu/n^2)\sim\frac1\pi\int_0^\pi e^{\sqrt{2\nu}\cos\theta}\,d\theta=I_0(\sqrt{2\nu}).}
$$
This <real integral representation of the modified Bessel function I0> gives one at $\nu=0$, agreeing with the endpoint value. Applying the same endpoint <Laplace method> to the integral for $I_0(z)$ gives $I_0(z)\sim e^z/\sqrt{2\pi z}$. Thus, as $\nu\to\infty$, the inner result is $e^{\sqrt{2\nu}}/\sqrt{2\pi\sqrt{2\nu}}$. The outer result has precisely this limit in the overlap $1\ll\nu\ll n$. This proves both matches rather than extending the fixed-$x$ formula directly to $x=1$.
For the derivatives, the <power series> $I_0(\sqrt{2\nu})=1+\nu/2+\nu^2/16+\cdots$ suggests their leading orders. To justify differentiation, expand the original integral itself near $x=1$. Its first three even cosine moments give
$$
P_n(x)=x^n+\frac{n(n-1)}4x^{n-2}(x^2-1)
+\frac{n(n-1)(n-2)(n-3)}{64}x^{n-4}(x^2-1)^2+O((x-1)^3).
$$
Odd cosine moments vanish, and higher even moments start at cubic order. Expanding at $x=1$ yields
$$
\boxed{P_n'(1)=\frac{n(n+1)}2\sim\frac{n^2}2,\qquad
P_n''(1)=\frac{(n-1)n(n+1)(n+2)}8\sim\frac{n^4}8.}
$$
For large $n$, the graph starts at $(1,1)$ with a very large positive slope and positive <curvature>. It rises through the $n^{-2}$ <boundary layer> and then follows the rapidly increasing outer <asymptotic expansion>. The sketch below shows both scales.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-82-legendre-asymptotics.png]
{title=Large-degree Legendre functions and their n-squared endpoint scaling}
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