Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition

Question

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?

Answer 1

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

Answer 2

Dividing out an $x^2$, you have to bound

$$\left | \frac{4-\frac{7}{x^2}}{2x-\frac{5}{x^2}} \right |.$$

Suppose $x>\sqrt{5}$. Then $4-\frac{7}{x^2}>0$, so $\left | 4 - \frac{7}{x^2} \right | = 4 - \frac{7}{x^2}<4.$ Additionally $\left | 2x-\frac{5}{x^2} \right | = 2x - \frac{5}{x^2} > 2x-1.$ So the whole thing is less than

$$\frac{4}{2x-1}.$$

Additionally we will have $x>1$ here, so $2x-1>x$. Hence the whole thing is less than

$$\frac{4}{x}.$$

Can you take it from here?

The hard part was finding the $\sqrt{5}$. The idea is ultimately that (after factoring the $x^2$) the numerator is very nearly $4$ and the denominator is very nearly $2x$, provided $x$ is large. We just have to choose $x$ large enough to take advantage of that, and be sure to replace terms with things that make the quantity smaller. In the numerator this was easy, because that term makes the entire expression smaller. In the denominator it was harder, because the term we wanted to remove was actually making the entire expression bigger. So we had to tweak the "good" term to balance out removing the "bad" term.