Monotonicity and asymptotics of the first eigenvalue of Laplace operator in annuli Trying to prove something else, I am lead to the following. Let $A_R$ be the annulus with radii $1$ and $R$, that is
$$ A_R := \{ x \in \mathbb{R}^n : 1 < |x| < R \} , $$
and consider de first Dirichlet eigenvalue of the Laplacian
$$ \lambda_R := \inf_{\substack{\phi \in H_0^1(A_R) \\ \phi \neq 0}} \frac{\int_{A_R} |\nabla \phi|^2 \; dx}{\int_{A_R} |\phi|^2 \;dx} ,$$
where $H_0^1(A_R)$ is the usual Sobolev space. My intuition tells me that we should have the following asymptotic behaviour:
$$ \lim_{R \to 1+} \lambda_R = +\infty \quad \text{and} \quad \lim_{R \to +\infty} \lambda_R = 0 , $$
although I'm unable to prove nor disprove the statements above.
Because up to extension by $0$ we have the inclusion $H_0^1(A_R) \subset H_0^1(A_S)$ for $R<S$, it is easy to se that
$$ \lambda_R > \lambda_S \quad \text{for} \quad R < S , $$
but that's the best I can conclude.
I would thankfully appreciate any ideas or comments. 
 A: It can be proved in many ways, but, perhaps, one of the simplest (and constructive) would be as follows.
Since we are in radial settings, using the spherical coordinates we can equivalently define $\lambda_R$ as
$$
\lambda_R = \inf \frac{\int_1^R r^{n-1} |u'|^2 \, dr}{\int_1^R r^{n-1} |u|^2 \, dr}.
$$
Since $1 \leq r^{n-1} \leq R^{n-1}$ for any $r \in (1,R)$, we see that
$$
\lambda_R = \inf \frac{\int_1^R r^{n-1} |u'|^2 \, dr}{\int_1^R r^{n-1} |u|^2 \, dr} \geq \frac{1}{R^{n-1}} \inf \frac{\int_1^R |u'|^2 \, dr}{\int_1^R |u|^2 \, dr}
= \frac{1}{R^{n-1}}\frac{\pi^2}{(R-1)^2}.
$$
(The last equality follows from the expression for the first eigenvalue of 1D problem.)
Thus, we see that if $R \to 1+$, then $\lambda_R \to +\infty$.
Let us show now that $\lambda_R \to 0$ when $R \to +\infty$.
As you have written, we have the so-called domain monotonicity: if $\Omega_1 \subsetneq \Omega_2$, then $\lambda(\Omega_1) > \lambda(\Omega_2)$.
Let us use it. Clearly, we have
$$
B_{\frac{R-1}{2}} \subset A_R,
$$
where $B_{\frac{R-1}{2}}$ is a ball of radius $\frac{R-1}{2}$ centred in an appropriate point. (Notice that $R-1$ is the widht of $A_R$.)
So, by the domain monotonicity, we have
$$
\lambda(B_{\frac{R-1}{2}}) > \lambda(A_R).
$$
On the other hand, the scaling property of $\lambda$ says
$$
\lambda(B_s) = \frac{\lambda(B_1)}{s^2},
$$
and hence we have
$$
\frac{4 \lambda(B_1)}{(R-1)^2} > \lambda(A_R).
$$
Since $\lambda(B_1)$ is just a constant which does not depend on $R$, we conclude that $\lambda_R \to 0$ whenever $R \to +\infty$.
