Interchange of limit operator and $\ln$ function. $$\lim_{n\to \infty}\ln\left(\frac{1+a^2n^2}{1+n^2}\right)$$
Can someone help evaluate that for me?

Hint: Since $\ln$ is a continuous function in $(0,+\infty)$ you can interchange the $\lim$ operator with $\ln$. Thus $$\lim_{n \to \infty} \ln\left(\frac{1+a^2n^2}{1+n^2}\right)=\ln\left(\lim_{n\to \infty}\frac{1+a^2n^2}{1+n^2}\right)=\ln\left(\lim_{n\to \infty}\frac{1/n^2+a^2}{1/n^2+1}\right)=\ln(a^2)$$
