I'm stuck on the epsilon-delta proof of a limit. Namely, Consider the function $f(x) = 2x^2 + 3x$ for $x \in \mathbb{R}$ and $0 < x < 4$. Prove that $\lim_{x \to 2} f(x) = 14$.
By the definition of a limit: $$\forall \epsilon > 0 \, \exists \delta > 0 \, \forall x(0 < |x-2|<\delta \Longrightarrow |2x^2 + 3x -14|< \epsilon)$$ So, to find $\delta$ in terms of $\epsilon$: $ \,\,\,\,\,\,\,\,\,\,\,\, |2x^2 + 3x -14| < \epsilon \\ \Longleftrightarrow |(2x+7)(x-2)|<\epsilon \\ \Longleftrightarrow|2x+7|\cdot|x-2|<\epsilon$
Now, since, I'm given that $0 < x < 4$. I could deduce, $|x-2| <2$, and $|2x+7| < 7$. But, I'm unsure how to use these inequalities to get $\delta$ in terms of $\epsilon$.
Since $0<x<4$ one has that $|2x+7|<15$ and since $|x-2|<\delta$ choosing $\delta = \frac{\varepsilon}{15}$ you get $$|2x+7||x-2|<15\cdot \delta=15\cdot \frac{\varepsilon}{15}=\varepsilon$$ as you desired to prove.