Given $\lim _ { x \rightarrow - 4 } ( 2 x + 6 ) = - 2$ find a $\delta 0$ in terms of an $\varepsilon 0$ so that if $0

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Given $\lim _ { x \rightarrow - 4 } ( 2 x + 6 ) = - 2$ find a $\delta > 0$ in terms of an $\varepsilon > 0$ so that if $0 < | x - c | < \delta$ then $| f ( x ) - L | < \varepsilon$ A) /3 B) /2 C) 1 D) None of these

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The Answer of Given $\lim _ { x \rightarrow - 4 } ( 2 x + 6 ) = - 2$ find a $\delta > 0$ in terms of an $\varepsilon > 0$ so that if $0 < | x - c | < \delta$ then $| f ( x ) - L | < \varepsilon$ A) /3 B) /2 C) 1 D) None of these

Given lim⁡x→−4(2x+6)=−2\lim _ { x \rightarrow - 4 } ( 2 x + 6 ) = - 2limx→−4​(2x+6)=−2

Given $\lim _ { x \rightarrow - 4 } ( 2 x + 6 ) = - 2$