Limit of $\frac{9n}{n+7}$ I wonder if there are "better"/faster ways to determine the limit without polynomial-division with the limit being -9.
The original exercise is $g_n := \frac{3}{n+7} \cdot (\sum_{k=0}^nk)-\frac{3}{2}n$
Then at some point you get to this term and although intuitively it is obvious that the limit is -9 I struggle to construct a null-sequence out of this too finally "prove" it
$\frac{-18n}{2n+14}$ = $\frac{-9n}{n+7}$ = $\frac{-9}{1} \cdot \frac{n}{n+7}$
With polynomial-divison you easily get to the following, but I never learned it and im sure there is a good way without it
$-9+\frac{126}{2n+14}$ 
Thanks in advance!

$$\frac{9n}{n+7}=\frac{9}{1+\frac{7}{n}}$$
