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Quiz 5
Prove $$\displaystyle\lim_{x \rightarrow 2} 3x^2+7x-14=12.$$ Proof: Let $\epsilon > 0$ and choose $0 < \delta < \min \left\{ 1, \dfrac{\epsilon}{22} \right\}$. Then if $|x-2| < \delta$, compute $$\begin{array}{ll} \left| (3x^2+7x-14) - 12 \right| &= \left| 3x^2+7x-26 \right| \\ &= \left| (3x^2-12) + (7x-14) \right| \\ &=\left| 3(x^2-4) + 7(x-2) \right| \\ &=\left| 3(x-2)\underbrace{(x+2)}_{=(x-2)+4} + 7(x-2) \right| \\ &=\left| 3(x-2)^2 + 12 (x-2) + 7(x-2) \right| \\ &=\left| 3\delta^2 + 19\delta \right| \\ &=\delta \left| 3\delta + 19 \right| \\ &\stackrel{\delta < 1}{<} 22\delta \\ &< \left( \dfrac{\epsilon}{22} \right) (22) \\ &= \epsilon. \blacksquare \end{array}$$