Prove the statement using the ε and δ definition of a limit:
$$\lim_{x\to3} \quad \frac{x}{(3x-10)} = -3 $$
Let $\epsilon>0$ be given. Find $\delta>0$ which depends on $\epsilon$ so that if $0<|x-3|<\delta$ and solve for $|x-3|$.
Then:
$$|f(x)-3|<\epsilon \text{ iff } |\frac{x}{3x-10} - 3|<\epsilon$$
$$\text{iff } |\frac{x}{3x-10} - \frac{3(3x-10)}{3x-10}| <\epsilon$$
$$\text{iff } |\frac{x-(9x-30)}{3x-10}| <\epsilon$$
$$\text{iff } |\frac{30-8x}{3x-10}|<\epsilon$$
$$\text{iff } |\frac{(-2)(15-4x)}{3x-10}| <\epsilon$$
$$\text{iff } 2 \frac{|15-4x|}{|3x-10|} <\epsilon$$
Arbitarily assume that $δ < 1$.
Then $|x-3|<\delta< 1$ implies that $-1<x-3<1$ and $2<x<4$.
However, range of $x$ values is not appropriate since $f(x)= \frac{x}{3x-10}$ is not defined at $10/3$. We must simply pick $\delta$ small enough to avoid $x=10/3$. For example, assume that $\delta< 1/4$. Then $|x-3|<δ< 1/4$ implies that $-1/4<x-3<1/4$ and $11/4<x<13/4$ so that $2<|3x-10|<4$ and $1/4<\frac{1}{|3x-10|}<1/2$
It follows that
$$\frac{2|15-4x|}{|3x-10|} < 2|x-3|\frac{1}{2} < \epsilon\text{ (this is where I think I went wrong)}$$
$$\text{iff } 2|x-3| \frac{1}{2} <\epsilon$$
$$\text{iff } |x-3| <\epsilon$$
Now choose $\delta = \min (1/4, \epsilon)$
Then I am not sure where to go from there...
Your very first mistake is that as we want to prove $\lim_{x\to 3}\frac x{3x -10}= \color{red}- 3$ we must find $\delta$ based on
$|f(x) - (\color{red}-3)| < \epsilon$ or
$|f(x) + 3| < \epsilon$
so $|\frac {x}{3x-10} + 3| < \epsilon$
$|\frac {x + 3(3x-10)}{3x-10}| = |\frac {10x -30}{3x-10}|<\epsilon$.
At this point I do something a little different
$|\frac {10x -30}{3x-10}|=|\frac {10x -\frac {10}3*10 + \frac {10}3*10 -30}{3x-10}|=|\frac {10}3 +\frac{3\frac 13}{3x-10}|=|\frac {10}3 +\frac {10}{9x-30}|<\epsilon$
$-\epsilon < \frac {10}3 +\frac {10}{9x-30}< \epsilon$
$-3\frac 13 -\epsilon < \frac {10}{9x-30}< -3\frac 13 +\epsilon$. If we assume $0 < \epsilon < 3\frac 13$ then all those terms are negative so
$0 < 3\frac 13 - \epsilon < \frac {10}{30-9x} < 3\frac 13 + \epsilon$ and so
$10 -3\epsilon < \frac {10}{10-3x} < 10 + 3\epsilon$
$\frac 1{10-3\epsilon} > \frac {10-3x}{10} > \frac 1{10+3\epsilon}$
$\frac {10}{10-\epsilon} > 10 -3x > \frac {10}{10+3\epsilon}$
$-10 + \frac {10}{10-\epsilon} > -3x> -10 + \frac {10}{10+3\epsilon}$
$\frac {10 - \frac {10}{10-\epsilon}}3 < x < \frac {10 - \frac {10}{10+3\epsilon}}3$
Okay... we can't put it off any longer:
we want $\frac {10 - \frac {10}{10-\epsilon}}3 < 3-\delta < 3-x < 3+ \delta < \frac {10 - \frac {10}{10+3\epsilon}}3$
So we need $\frac {10 - \frac {10}{10-\epsilon}}3 < 3-\delta$ so we solve for $\delta$. $\delta < 3-\frac {10 - \frac {10}{10-\epsilon}}3= \frac {10}{30-3\epsilon}-\frac 13$.
And we need $3+ \delta < \frac {10+\frac {10}{10+3\epsilon} }3$ or $\delta < \frac 13 -\frac{10}{30 +3\epsilon}$.
So if we set $\delta = \min (\frac {10}{30-3\epsilon}-\frac 13, \frac 13 -\frac{10}{30 +3\epsilon})$ and we are done.