# `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...

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`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.`

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