Related to the example on the rectangle in the book Strauss W.A. Partial differential equations - an introduction (Wiley, $2008$, $2$nd Ed.) page $326$, is there anyone could explain to me how is it possible to go from $\frac{\lambda_n a b}{4 \pi} - C \sqrt{\lambda_n} \leq n \leq \frac{\lambda_n a b}{4 \pi}$ to $\lim_{n \to \infty} \frac{\lambda_n}{n}=\frac{4 \pi}{ab}$ ?
The right inequality implies $$ \frac{\lambda_n}{n}\geq \frac{4\pi}{ab}. $$ We use that in the left inequality (for the square root term only), $$ n\geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\lambda_n} \geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\frac{4\pi n}{ab}}. $$ Rearranging, $$ \frac{\lambda_n}{n}\leq\frac{4\pi}{ab}+C\Bigl(\frac{4\pi}{ab}\Bigr)^{3/2}\frac{1}{\sqrt{n}}. $$ Then apply the sandwich lemma for sequences.