Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition I am having difficulties manipulating the problem so that I can find a $N$ value to choose.
Suppose $x > N$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| \leq \frac{4x^2}{|2x^3 - 5|} + \frac{7}{|2x^3 - 5|} $$
This is basically as far as I gone which I know is correct. I am not sure what would be the next step. Any help/hints would be appreciated!
Edit:
I found an interesting way to approach it, not sure if it is right.
Assume $x > 2$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| < \frac{4x^2}{2x^3} = \frac{2}{x} < \frac{2}{N} < \epsilon$$
That means I can choose $N = \max\left\lbrace 2, \frac{2}{\epsilon}\right\rbrace$
Is that correct?

Not quite, you made your denominator larger there, so that step is not obvious. That is to say, $2x^3-5\le 2x^3\iff {1\over 2x^3-5}\ge {1\over 2x^3}$, so you cannot conclude that ${4x^2-7\over 2x^3-5}\le {4x^2\over 2x^3}$. Instead I recommend the following:
For $x>2$ we have $4x^2>4x^2-7$ and $2x^3-5\ge 2x^3-x^3$, i.e. ${1\over 2x^3-5}\le {1\over 2x^3-x^3}$, since both quantities are positive. So we have

$$\left|{4x^2-7\over 2x^3-5}\right|\le {4x^3\over x^3}={4\over x}$$

From this we can let $N=\max\left\lbrace 2,\left[{4\over\epsilon}\right]+1\right\rbrace$.
