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 $\displaystyle \epsilon >0$ there exsists an $n_{0}$ such that

$\displaystyle \left|{\frac {3n^{2}-4n}{n^{2}+5n+2}}-3\right|<\epsilon$

for all $n\geq n_{0}$. Let $\displaystyle \epsilon >0$. Simplifying the left hand side above yields:

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

Now, choose $n_{0}$ large enough so that $\displaystyle {\frac {19}{n_{0}}}<\epsilon$ which can be done by choosing $n_{0}$ larger than $\displaystyle {\frac {19}{\epsilon }}$. Then, for each $n\geq n_{0}$, we have that

${\begin{aligned}\left|{\frac {3n^{2}-4n}{n^{2}+5n+2}}-3\right|&<{\frac {19}{n}}<\epsilon \end{aligned}}$

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