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# Prove the statement using the $\varepsilon$, $\delta$ definition of a limit. $\displaystyle \lim_{x \to 4}\frac{x^2 - 2x - 8}{x - 4} = 6$

## Given $e>0,$ we need $\delta>0$ such that if $0<|x-4|<\delta$, then $\left|\frac{x^{2}-2 x-8}{x-4}-6\right|<\varepsilon \Leftrightarrow$$\left|\frac{(x-4)(x+2)}{x-4}-6\right|<\epsilon \Leftrightarrow|x+2-6|<\varepsilon \quad[x \neq 4] \Leftrightarrow|x-4|<\varepsilon,$ So choose $\delta=\epsilon,$ Then$0<|x-4|<\delta \Rightarrow|x-4|<\varepsilon \Rightarrow|x+2-6|<\varepsilon \Rightarrow\left|\frac{(x-4)(x+2)}{x-4}-6\right|<\varepsilon[x \neq 4] \Rightarrow$$\left|\frac{x^{2}-2 x-8}{x-4}-6\right| < t .$ By the definition of a limit, $\lim _{x \rightarrow 4} \frac{x^{2}-2 x-8}{x-4}=6$

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Missouri State University

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University of Michigan - Ann Arbor

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Idaho State University

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