Science:Math Exam Resources/Courses/MATH220/December 2009/Question 07 (b)/Solution 1

We show that the sequence converges to 3. To see this, it suffices to show that for any $ϵ>0$ there exsists an $n_0$ such that

$|\frac{3n^2-4n}{n^2+5n+2}-3|<ϵ$

for all $n\ge n_0$. Let $ϵ>0$. Simplifying the left hand side above yields:

$\begin{array}{rl}|\frac{3n^2-4n}{n^2+5n+2}-3|& =|\frac{3n^2-4n}{n^2+5n+2}-\frac{3(n^2+5n+2)}{n^2+5n+2}|\\ & =\left|\frac{-19n-6}{n^2+5n+2}\right|\\ & =\frac{19n+6}{n^2+5n+2}\\ & <\frac{19n+19}{n^2+n}\\ & =\frac{19(n+1)}{n(n+1)}\\ & =\frac{19}{n}\end{array}$

Now, choose $n_0$ large enough so that $\frac{19}{n_0}<ϵ$ which can be done by choosing $n_0$ larger than $\frac{19}{ϵ}$. Then, for each $n\ge n_0$, we have that

$\begin{array}{rl}|\frac{3n^2-4n}{n^2+5n+2}-3|& <\frac{19}{n}<ϵ\end{array}$

which shows that our original sequence converges to $3$.