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Prove using epsilon,delta definition of limit.

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lim (5x + 8) = 3 as x -> -1

asked Sep 5, 2014 in PRECALCULUS by anonymous

1 Answer

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We must show that for each ε > 0, there exists a δ > 0 such that |(5x + 8) - 3| < ε whenever 0 < |x -(-1)| < δ. Because our choice of δ on ε, we need establish a connection between the absolute values |(5x + 8) - 3| and |x -(-1)|.

This is usually done by starting from the consequent :

|(5x + 8) - 3| = |(5x + 5|  = |5(x + 1)| = |5(x + 1)| = 5|x + 1|

So, for a given ε > 0 we can choose δ = ε/5. This choice works because

0 < |x -(-1)| < δ = ε/5  ⇒ |(5x + 8) - 3| = 5|x + 1| < 5(ε/5) = ε.

answered Sep 5, 2014 by casacop Expert

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