Give an $\varepsilon-\delta$ proof using the “$\delta\leq 1 $ trick” to prove that $\lim_{x\rightarrow c} (2x^ 2 − 3x + 4) = 2c ^2 − 3c + 4$.
My attempt: Let $ε>0$ and there exist $\delta>0$ such that whenever $|x-c|<δ$, $$|2x^ 2 − 3x + 4-(2c ^2 − 3c + 4)|<ε$$
$$\implies |2x^2-2c^2-3x+3c|<ε$$
$$\implies 2|x^2-c^2|-3|x-c|<ε$$
$$\implies 2|x-c||x+c|-3|x-c|<ε$$
$$\implies |x-c||2x+2c-3|<ε$$
I don't see where to go from here.
Note that $$|x+c|\leq |x-c|+|2c|$$ which means $$2|x-c||x+c|+3|x-c|\leq 2|x-c|^2+4|c||x-c|+3|x-c|=|x-c|(2|x-c|+4|c|+3)$$
Choose $$\delta:=\min\left\{\frac{\epsilon}{8|c|+6}, 2|c|+\frac{3}{2}\right\}$$ Then for $|x-c|<\delta$, we have $$|x-c|(2|x-c|+4|c|+3)<\delta(2\delta+4|c|+3)\leq\frac{\varepsilon}{8|c|+6}\left(2\left(2|c|+\frac{3}{2}\right)+4|c|+3\right)=\epsilon$$
which gives $$|(2x^2-3x+4)-(2c^2-3c+4)|\leq |x-c|(2|x-c|+4|c|+3)<\epsilon$$