This is the same as finding regular limits, but only using integers, right?
The problem is: a_n=[ln(2+e^n)]/3n
How do I solve it?
Or merely note that $\displaystyle \frac{\ln\left(2+e^n\right)}{3n}=\frac{\ln\left(e^ n\left(\frac{2}{e^n}+1\right)\right)}{3n}=\frac{\l n\left(e^n\right)}{3n}+\frac{\ln\left(\frac{2}{e^n }+1\right)}{n}=\frac{1}{3}+\frac{\ln\left(\frac{2} {e^n}+1\right)}{n}$. The second term in a senes "doubly" tends to zero. The numerator tends towards $\displaystyle \ln\left(1\right)=0$ and the denominator is getting unboundedly large.