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Quick Homework Help
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I would really appreciate a thorough explanation of the following limit problem...thanks in advance.

JDavid

by JDavid at June 07, 2011

f of x = 1/(x-1)
Find delta such that if 0<absolute value of (x-2)<delta then absolute value of (f of x minus 1) < 0.01

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JDavid -Prove: Given ε>0:|f(x) - 1| < ε = 0.01|x-2| < 0.01|x-1|Since we require |x-2| be small, let's assume:|x - 2| < 0.5 or 1.5 < x < 2.5. So, if x > 1.5:Let delta < 0.01|x-1| < 0.01|1.5 - 1| = 0.01|0.5| = .005Rigorous Proof:Hence, for 0 < |x - 2| < delta = .005|x - 2| < .005 = .01|0.5||x - 2| ÷ |0.5| < 0.01|x - 2| ÷ |x - 1| < 0.01 = ε| 1/(x-1) - 1| < 0.01 = ε| f(x) - L | < 0.01 = εHope that helps

Steve204 Steve204 June 08, 2011

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