We have to count following integral: $$ \lim_{n\to\infty}\int_A\frac{\ln(n+y^3)-\ln n}{\sin(x/n)}d\lambda_2 $$ where $A=\{(x,y)\in\mathbb{R}^2_+:1<xy<4,1<y/x<4\}$
Now in order to find integrable bound of above function, it may be useful to substitute variables: $\phi(x,y)=(s,st)$ and $\det\phi=s$, so now we integrate over set $\{(s,t)\in\mathbb{R}^2_+:1<s^2t<4,1<t<4\}$
$$ \int_A\frac{\ln(n+y^3)-\ln n}{\sin(x/n)}d\lambda_2=\int_A\frac{\ln(\frac{n+y^3}{n})}{\sin(x/n)}d\lambda_2=\int\frac{\ln(1+(st)^3/n)s}{\sin(s/n)}dsdt $$ and $$\frac{\ln(1+(st)^3/n)s}{\sin(s/n)}<\frac{\ln(1+(4s)^3/{n})s}{\sin(s/n)}<(4s)^3$$which is integrable over bounded area.
And now I can say $$\lim\int\frac{\ln(1+(st)^3/n)s}{\sin(s/n)}dsdt=\int(st)^3dsdt$$
I have doubts, because I could just force this move without substitution. I could take $$\int_A\frac{\ln(1+\frac{y^3}{n})}{\sin(x/n)}d\lambda_2=\int_A\frac{\ln(1+\frac{y^3}{n})}{\sin(x/n)}\frac{x/n}{y^3/n}\frac{y^3/n}{x/n}d\lambda_2\to\int_A(y^3/x) dxdy$$ which is obviously different result under above substitution.
You can move the limit inside the integral as long as you can show that the inside function is being dominated by a integrable function $g$.
This is the result of the Dominated convergence theorem.