For fixed $x>0$,
$$
K_\nu(x)\sim\left(\frac\pi{2\nu}\right)^{1/2}
\left(\frac{2\nu}{ex}\right)^\nu
$$
as $\nu\to+\infty$. Applying <Laplace's method> to the <integral representation of the modified Bessel function of the second kind> proves the formula.

Large-order asymptotic of the modified Bessel function of the second kind

Use the real <integral representation of the modified Bessel function of the second kind>
$$
K_\nu(x)=\frac12\int_{-\infty}^{\infty}\exp\{-x\cosh t+\nu t\}\,dt
=\frac12\int_{-\infty}^{\infty}e^{\nu h(t)}\,dt,
\qquad h(t)=t-\frac{x}{\nu}\cosh t.
$$
The unique <stationary point> is the strict maximum
$$
t_0=\operatorname{arsinh}\frac\nu x,
\qquad
\nu h(t_0)=\nu\operatorname{arsinh}\frac\nu x-\sqrt{\nu^2+x^2},
\qquad
\nu h''(t_0)=-\sqrt{\nu^2+x^2}.
$$
The <Laplace's method>[Laplace approximation] is consequently
$$
K_\nu(x)\sim
\frac12\left(\frac{2\pi}{\sqrt{\nu^2+x^2}}\right)^{1/2}
\exp\left\{\nu\operatorname{arsinh}\frac\nu x-\sqrt{\nu^2+x^2}\right\}.
$$
For fixed $x>0$, as $\nu\to+\infty$,
$$
\operatorname{arsinh}\frac\nu x=\log\frac{2\nu}{x}+O(\nu^{-2}),
\qquad
\sqrt{\nu^2+x^2}=\nu+O(\nu^{-1}).
$$
Substitution gives the required <large-order asymptotic of the modified Bessel function of the second kind>:
$$
\boxed{K_\nu(x)\sim\left(\frac{\pi}{2\nu}\right)^{1/2}
\left(\frac{2\nu}{ex}\right)^\nu.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 31B / b / Solution

For real $x>0$ and $\nu$, one useful representation is
$$
K_\nu(x)=\frac12\int_{-\infty}^{\infty}e^{-x\cosh t+\nu t}\,dt.
$$
It turns large-order asymptotics into a <Laplace integral> with a moving maximum.

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 Incoming links: Integral representation of the modified Bessel function of the second kind