Limits for sequences: Prove that $\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$

Question

Prove that $$\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$$

Note, this is a sequence.

I know that a sequence $(c_n)_{n=1}^{\infty}$ converges to a finite value L if for every $\epsilon > 0$, there exists some $N>0$ so that for all $n$ it holds that: $$n > N \implies |c_n - L| < \epsilon$$.

I've learnt this in school: $|c_n - L| < \epsilon$ is the distance between each sequence term and the epsilon-value. If we can write $n$ in terms of epsilon, then for any epsilon, we can easily 'choose' an N-value, so that the implication above holds.

I'm afraid I haven't gotten further than this... :

$$\left|\frac{2n^2 - 3n + 5}{9 - n^2} - (-2)\right| < \epsilon \iff \left|\frac{23 - 3n}{9 - n^2}\right| < \epsilon$$

I just can't seem to reduce $n$ in a way that would help me solve this inequality... I don't know if there is some algebraic trick that is going over my head or if there is some "theoretical trick" which I could use to simplify. Please, a hint would be greatly appreciated.

Answer 1

The quick way:

$\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$

Divide to and bottom by $n^2$

$\lim_{n \rightarrow \infty} \frac{2 - \frac {3}n + \frac {5}{n^2}}{- 1 + \frac {9}{n^2}} = -2$

And when $n$ gets to be large the $\frac {1}n$ terms become small. The $\epsilon-\delta$ definition is not necessry.

But if that is the direction you want to go.

There exists an $N$ such that $n>N$ implies $\left|\frac {3n-23}{n^2-9}\right| < \epsilon$

let $N = \max (\frac {3}{\epsilon},4)$

Answer 2

Hint: prove that $\frac{23-3n}{9-n^2}\to 0$.

Answer 3

You want so show that $$\left|\frac{23 - 3n}{9 - n^2}\right| < \epsilon$$ First when $n$ is large enough ($n>8$) $$\left|\frac{23 - 3n}{9 - n^2}\right| =\frac{3n-23}{n^2-9}<\frac{3n}{n^2-9} $$ Also, if $n$ is this large, we have $n^2-9>\left(\frac{n}{2}\right)^2$ so $$\frac{3n}{n^2-9}<\frac{12n}{n^2}=\frac{12}{n}$$

Therefore, it's enough to take $$n > \max\left(8,\frac{12}{\epsilon}\right)$$

NOTE

I don't recommend doing it this way. You should just show the fraction goes to $0$ which guarantees the existence of such an $N$. Doing it this way is more tedious, no more rigorous, and in my view, adds nothing to understanding.

Answer 4

If we take $n > 3+\frac6\varepsilon$ we have $ \frac{3}{n-3} < \frac{\varepsilon}2$, and if we take $n > \sqrt{\frac{28}\varepsilon + 9}$ we have $ \frac{14}{n^2-9} < \frac{\varepsilon}2$.

Therefore for $n \ge \max\left\{4, 3+\frac6\varepsilon, \sqrt{\frac{28}\varepsilon + 9}\right\}$ we have

\begin{align} \left|\frac{2n^2-3n+5}{9-n^2}+2\right| &= \frac{|23-3n|}{|9-n^2|}\\ &\le \frac{3n+23}{(n+3)(n-3)}\\ &= \frac{3(n+3)+14}{(n+3)(n-3)}\\ &= \frac{3}{n-3} + \frac{14}{n^2-9}\\ &< \frac{\varepsilon}2 + \frac{\varepsilon}2\\ &= \varepsilon \end{align}