Interchange of limit operator and $\ln$ function.

Question

$$\lim_{n\to \infty}\ln\left(\frac{1+a^2n^2}{1+n^2}\right)$$

Can someone help evaluate that for me?

Answer 1

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

Answer 2

As Stef said, $\ln$ is continuous, so you have : $$ \lim_{n\to \infty}\ln\left(\frac{1+a^2n^2}{1+n^2}\right) = \ln \left(\lim_{n\to \infty}\left(\frac{1+a^2n^2}{1+n^2}\right)\right)$$

Then : $$\lim_{n\to \infty}\left(\frac{1+a^2n^2}{1+n^2}\right) = \lim_{n\to \infty}\left(\frac{a^2n^2}{n^2}\right) = a^2 $$

And finally : $$ \lim_{n\to \infty}\ln\left(\frac{1+a^2n^2}{1+n^2}\right) = \ln \left(\lim_{n\to \infty}\left(\frac{1+a^2n^2}{1+n^2}\right)\right) = \ln(a^2)$$

This is valid if you suppose $a^2 \neq 0$. If $a = 0$, you have : $$\lim_{n\to \infty}\ln\left(\frac{1+a^2n^2}{1+n^2}\right) = \lim_{n\to \infty}\ln\left(\frac{1}{1+n^2}\right) = -\infty$$