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?

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- Nov 23rd 2009, 05:45 PMschnekSequence Limits
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? - Nov 23rd 2009, 05:51 PMapcalculus
- Nov 23rd 2009, 06:32 PMDrexel28
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.