The question is as follows:
The definition for ${ \lim_{x\to\infty} } = L$ is $\forall \epsilon \gt0 , \exists M\gt0$ such that if $x > M$ then $\left| f(x) - L \right| < \epsilon$.
Use this definition to prove that $\displaystyle\lim_{x \to \infty} \frac{4x^3+7x+19}{2x^3+3}=2$.
I'm able to do the first step, more or less just rationalizing to $\displaystyle \left| \frac{7x+13}{2x^3+3} \right|< \epsilon $. Then I'm lost. Anyone able to help with a solution is appreciated.
Note that if $x>13$, then $7x+13<8x$, and that $2x^3+3>2x^3$. Therefore$$\left|\frac{7x+13}{2x^3+3}\right|=\frac{7x+13}{2x^3+3}<\frac{8x}{2x^3}=\frac4{x^2}.$$So, if $x>\max\left\{13,\sqrt{\frac4\varepsilon}\right\}$, then$$\left|\frac{7x+13}{2x^3+3}\right|<\frac4{x^2}<\varepsilon.$$