Question 1027615: Prove that the limit as x approaches 5 of f(x) = ax+b is 5a+b using delta-epsilon proof.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We start by determining a value for delta before starting the proof
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|f(x) - L| < epsilon, where L is the limit
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|ax+b - 5a-b| < epsilon
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|a||x-5| < epsilon
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|x-5| < epsilon/a
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|x-c| < delta
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Note that c=5, since x --> 5
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Therefore
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delta = epsilon/a
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Proof starts
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epsilon > 0 is given
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delta = epsilon/a
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epsilon > 0, then delta > 0
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0 < |x-c| < delta implies
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|x-5| < epsilon/a
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|ax-5a| < epsilon
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Therefore
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limit as x-->5 (ax+b) = (5a+b)
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Note that we added b to both sides of the -
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