Answer
Limit: $0.5$
Work Step by Step
We are given the sequence:
$a_n=\dfrac{3n}{1+6n}$
Determine the first 10 terms of the sequence:
$a_1=\dfrac{3(1)}{1+6(1)}=\dfrac{3}{7}\approx 0.42857$
$a_2=\dfrac{3(2)}{1+6(2)}=\dfrac{6}{13}\approx 0.46154$
$a_3=\dfrac{3(3)}{1+6(3)}=\dfrac{9}{19}\approx 0.47368$
$a_4=\dfrac{3(4)}{1+6(4)}=\dfrac{12}{25}\approx 0.48000$
$a_5=\dfrac{3(5)}{1+6(5)}=\dfrac{15}{31}\approx 0.48387$
$a_6=\dfrac{3(6)}{1+6(6)}=\dfrac{18}{37}\approx 0.48649$
$a_7=\dfrac{3(7)}{1+6(7)}=\dfrac{21}{43}\approx 0.48837$
$a_8=\dfrac{3(8)}{1+6(8)}=\dfrac{24}{49}\approx 0.48979$
$a_9=\dfrac{3(9)}{1+6(9)}=\dfrac{27}{55}\approx 0.49091$
$a_{10}=\dfrac{3(10)}{1+6(10)}=\dfrac{30}{61}\approx 0.49180$
The sequence appears to have the limit $0.5$.
Plot the graph:
