Limit of a rational sequence

Text

Compute $\displaystyle\lim_{n\to+\infty}\frac{3n^{2}+2n}{n^{2}-5}$.

Solution

It is an indeterminate form $\tfrac{\infty}{\infty}$. Divide numerator and denominator by the dominant power $n^{2}$: $\frac{3n^{2}+2n}{n^{2}-5}=\frac{3+\dfrac{2}{n}}{1-\dfrac{5}{n^{2}}}.$ As $n\to+\infty$ the terms $\tfrac{2}{n}$ and $\tfrac{5}{n^{2}}$ tend to $0$, so the limit is $\dfrac{3}{1}=3$. $\boxed{\,L=3\,}$