I want to see if my proof is correct and if my choice of $\delta$ makes sense.
Prove that $\lim_{x\to 1} \frac{x-1}{2x^2+x-3} = \frac{1}{5}$. Let $\delta = \min(1, \frac{15}{2} \epsilon)$ and suppose $0 < |x-1| < \delta$.
First simplifying $|\frac{x-1}{2x^2+x-3} - \frac{1}{5}| = \frac{2}{5(2x+3)} |x-1|$. Since $|x-1| < 1$ we have $\frac{1}{2x+3} < \frac{1}{3} $.
So $\frac{2}{5(2x+3)} |x-1| < \frac{2}{15} |x-1| < \frac{2}{15} \delta \le \frac{2}{15} \frac{15}{2} \epsilon = \epsilon$.
For further details $|x-1| < 1 \iff -1<x-1<1 \iff 0<x<2 \iff 0<2x<4 \iff 3<2x+3<7 \implies \frac{1}{2x+3} < \frac{1}{3}$