Is $0$ a limit point of this parametrized family of integrals?

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Let $n \ge 3$ be a fixed integer, and let $\lambda >0$ be a real number.

I am interested in the behaviour of the following integral as a function of $\lambda$:

$$E(\lambda)=\lambda^2 \int_{0}^{\infty} (\frac{1}{ 1+\lambda^2r^2 })^2 \cdot (\frac{1}{1+r^2})^{n-2}r^{n-1} dr$$

  1. I would like to know if $0$ is a limit point of the set $\{ E(\lambda) \, | \, \lambda >0\}$.

  2. In particular, is it true that $\lim_{\lambda \to 0} E(\lambda)=0$? What about $\lim_{\lambda \to \infty} E(\lambda)$?

  3. Finally, is there a direct way to see that $E(\lambda)$ is maximal for $\lambda=1$? (A naive idea would be to try and "differentiate under the integral sign" w.r.t $\lambda$, but this doesn't work for improper integrals).

Of course, the best result would be an explicit calculation of $E(\lambda)$, but that's a bonus.

Motivation:

$E(\lambda)$ is the Dirichlet energy of a map $\psi_{\lambda}:\mathbb{S}^n \to \mathbb{S}^n$;

$\psi_{\lambda}$ is obtained by using the stereographic projection, then dilating by $\lambda$ and then projecting back. $(\psi_{\lambda})_{\lambda >0}$ is a family of conformal diffeomorphisms homotopic to the identity ($\psi_1$). I am asking if its energy comes arbitrarily close to zero.

(In dimension $n=2$, the Dirichlet energy is conformal invariant, hence $E(\lambda)$ is independent of $\lambda$, as can also be verified by a direct calculation).