Math 125-103
Fall 2012
Quiz 6
Carter
Name
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√
According to my calculator, 2 − 3 = 0.26795 while 5 − 2 = 0.23607. Consider the function
y = x2 in a neighborhood of c = 2. Find a δ > 0 so that |x2 − 4| < 1 provided |x − 2| < δ. See the
blackboard for an illustration.
Solution: We want |x2 − 4| < 1. So
−1 < x2 − 4 < 1
or
3 < x2 < 5
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√
while x is sufficiently close to 2 (and therefore positive). Thus 3 < x < 5. Subtract 2 from all
sides of the inequalities:
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√
3 − 2 < x − 2 < 5 − 2.
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This will hold true if δ = 5 − 2 which is the smaller of the two delimiting quantities.