proving the limit of a function using the $\epsilon , \delta$ definition

Use the $\epsilon , \delta$ definition of a limit to verify that $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}=2$$

Let, $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}-2 = \frac{x^2+x -6x+4}{3x-2}= \frac{x^2-5x+4}{3x-2} = \frac{(x-4)(x-1)}{3x-2}$$

but $x \ne \frac{2}{3}$ so we need to find $\lvert x-1 \rvert<$ some number betwen 1 and $\frac{2}{3}$ say $\frac{5}{6}$

So $\lvert x-1 \rvert < \frac{5}{6}$ or $\frac{1}{6}< x< \frac{11}{6}$

We want the upper bound so we use $x< \frac{11}{6}$ and the upper bound is $\frac{(\frac{11}{6}-4)(\frac{11}{6}-1)}{3(\frac{11}{6})-2}$ but that doesn't make any sense cause then the upper bound would be negative. Should I have used the lower bound?

Continuing your analysis: If $|x-1|<{5\over 6}$ then $$\left|{x^2+x\over 3x-2}-2\right|=\left|{(x-4)(x-1)\over 3x-2}\right|\le {23/6\over 3/2}\cdot|x-1|={23\over 9}|x-1|,$$ because $|x-4|<23/6$ and $|3x-2|>3/2$ for $x\in(1/6,11/6)$. Now make your choice of $\delta>0$ (for a given $\epsilon>0$) taking care that it is no larger than $5/6$ so that the above inequality is in force.