Use the definition of a limit to prove this limit Use the definition of a limit to show that $\frac{3x-2}{x^2+1} \to \frac{4}{5}$ as $x\to 2$.

I am having trouble isolating the $(x-2)$ from the fraction to use as a delta please help!
$$\left|\frac{3x-2}{x^2+1}-\frac45\right|=\left|\frac{15x-10-4x^2-4}{5(x^2+1)}\right|=\left|\frac{4x^2-15x+14}{5(x^2+1)}\right|=\frac{\left|4(x-2)\left(x-\frac74\right)\right|}{5(x^2+1)}=$$
(taking $\,x\,$ very close to $\,2\,$, say $\,x\in\left(2-0.001,\,2+0.001\right)\,$ , we can assure that $\,x-\frac74&gt;0\,$ , so)

$$=\frac{4x-7}{5(x^2+1)}|x-2|\;\;\color{red}{(**)}$$

and now we can estimmate, for $\;x\;$ as above (no need to be specially gentle or subtle with this...):

$$\frac{4x-7}{5(x^2+1)}&lt;\frac2{5\cdot(1+1)}=\frac1{5}\;\;\color{blue}{(*)}$$

so given any $\epsilon&gt;0\;$ we can choose $\;\delta=5\epsilon\;$ , and then:

$$\color{green}{(*)}\;\;|x-2|&lt;\delta\implies\left|\frac{3x-2}{x^2+1}-\frac45\right|\stackrel{\color{red}{(**)}}=\frac{4x-7}{5(x^2+1)}|x-2|\stackrel{\color{green}{(*)+\color{blue}{(*)}}}&lt;\frac15\cdot5\epsilon=\epsilon$$
You can simply set x=2 and solve the equation. You can do this because the equation is defined at x=2. If it were not then you would have to determine what value the function is approaching (usually infinity or zero, but can be anything).