Find $\lim_{n \rightarrow \infty} a_n$ where $$ a_n=\Bigg(\frac{3n^2}{3n^2-1}\Bigg)^{2n^2+3}$$
How to even start with an exponent problem like this ?
Find $\lim_{n \rightarrow \infty} a_n$ where $$ a_n=\Bigg(\frac{3n^2}{3n^2-1}\Bigg)^{2n^2+3}$$
How to even start with an exponent problem like this ?
Hint : Take logarithm both side and then take $n \to \infty$