a_n = (10n^2+3n+7)/(2n^2-6) Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

lim_(n to infty)(10n^2+3n+7)/(2n^2-6)=
Divide both numerator and the denominator by n^2.
lim_(n to infty)((10n^2)/n^2+(3n)/n^2+7/n^2)/((2n^2)/n^2-6/n^2)=lim_(n to infty)(10+3/n+7/n^2)/(2-6/n^2)=
Since lim_(n to infty) alpha/n^beta=0, forallalpha in RR, forall beta>0 we have
(10+0+0)/(2-0)=10/2=5
As we can see the sequence converges to 5.
The image below shows first 150 terms of the sequence. We can see they are asymptotically approaching the red line whose equation is y=5. 
http://mathworld.wolfram.com/ConvergentSequence.html