proving limit of a function using $\epsilon - N$.

Question

I'm doing an assignment and I've come across the following problem:

Prove the following using $\epsilon - N$: $$\lim_\limits{x \to \infty} \frac{3n^2 -2n + 1}{2n^2 - 4} = \frac{3}{2}$$

I'm having trouble isolating the n value in terms of $\epsilon$. I've gotten as far as:

$$\left|\frac{3n^2 -2n + 1}{2n^2 - 4} - \frac{3}{2}\right| = \left|\frac{7 - 2n}{2n^2 - 4}\right| < \left|\frac{7 - 2n}{n^2 - 2}\right|$$

and I'm unable to progress from here. I'm struggling to think of a "nice/simple" function to serve as my upper bound such that I can determine $n$ in terms of $\epsilon$.

I'm just looking for someone to point me in the right direction.

Answer 1

Note that$$n\geqslant4\implies\left|\frac{7-2n}{2n^2-4}\right|=\frac{2n-7}{2n^2-4}<\frac{2n}{2n^2-4}=\frac n{n^2-2}.$$But you have $n^2-2\geqslant\frac{n^2}2$, since this is equivalent to $n^2\geqslant4$. So,$$n\geqslant4\implies\left|\frac{7-2n}{2n^2-4}\right|<\frac n{n^2/2}=\frac2n.$$Can you take it from here?

Answer 2

For large $n$ we can assume $2n > 7$ and $2n^2 > 4$ so

$|\frac {7-2n}{2n^2-4}| = \left|\frac {2n -7}{2n^2 - 4}\right|=\left|\frac {2-\frac 7n}{2n-\frac 4n}\right|$.

Intuitively we know the $\frac kn$ ought to be insignificant and we we can argue that $2-\frac 7n < 2$ (assuming $n > 3\frac 12$) and $2n-\frac 4n > 2n-1 > 2n-n =n$ (if $n > 4$).

So $\left|\frac {2n-7}{2n^2-4}\right| < \left|\frac 2n\right| < \epsilon$ (assuming $n >\frac 2\epsilon$ and $n > 4$).

It's a bit crude but it works.

Perhaps a little less confusing. $2n - 7 < 2n -\frac 4n$ for $n > \frac 47$ so $\left|\frac {2n-7}{2n^4 -4}\right|< \left|\frac {2n-\frac 4n}{2n^2-4}\right| = \left|\frac 1n\right| < \epsilon$ (assuming $n > \frac 1\epsilon$ and $n > \frac 47$).

Or... perhaps more than we need:

$\frac {7-2n}{2n^2-4} = \frac {-1}{\frac {2n^2 -4}{2n-7}}=$

$-\frac 1{\frac {2n^2 -7n + 7n-4}{2n-7}}=-\frac 1{n + \frac {7n-4}{2n-7}}=$

$-\frac 1{n+\frac {7n-\frac 72\cdot7 + \frac 72 \cdot7-4}{2n-7}}=-\frac 1{n+\frac 72+\frac {45}{2n-7}}$

So $\left|\frac {7-2n}{2n^2-4}| =|\frac 1{n+\frac 72 +\frac {45}{2n-7}}\right|<\frac 1n<\epsilon$ (assuming $n> 3\frac 12$ and $n > \frac 1\epsilon$)

Answer 3

The ratio tends to $\dfrac2n$, so we consider the expression

$$\frac{2n-7}{n^2-2}\frac n2=\frac{2n^2-7n}{2n^2-4}$$ which is obviously in $[0,1]$ for $2n^2>7n>4$, i.e. $n>3$.