# Sequence Limits

• Nov 23rd 2009, 05:45 PM
schnek
Sequence 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 PM
apcalculus
Quote:

Originally Posted by schnek
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?

You could use L'Hopital's Rule here. Observe that both numerator and denominator go to infinity as n goes to infinity.

Then the ratio will approach the ratio of the derivatives:

$\frac{e^n}{(2+e^n)3} \rightarrow \frac{1}{3}$
• Nov 23rd 2009, 06:32 PM
Drexel28
Quote:

Originally Posted by schnek
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 $\frac{\ln\left(2+e^n\right)}{3n}=\frac{\ln\left(e^ n\left(\frac{2}{e^n}+1\right)\right)}{3n}=\frac{\l n\left(e^n\right)}{3n}+\frac{\ln\left(\frac{2}{e^n }+1\right)}{n}=\frac{1}{3}+\frac{\ln\left(\frac{2} {e^n}+1\right)}{n}$. The second term in a senes "doubly" tends to zero. The numerator tends towards $\ln\left(1\right)=0$ and the denominator is getting unboundedly large.