Name:Math 1A Quiz 2September 9, 2008GSI: Rob BayerYou have until 4:30 to complete this quiz. You must show your work.1. (3 pts) The graph of a function f(x) is pictured below. Find each of the following, or statethat they do not exist:(a) limx→0f(x) = 0(b) limx→−2+f(x) = -1(c) limx→1f(x) DNE(d) limx→1−f(x) =1(e) limx→2f(x) = 1(f) limx→−1f(x) = 02. (3 pts) Prove, using the − δ definition of a limit, that limx→−32x − 1 = −7Let > 0 be given. We want to find δ > 0 such that 0 < |x + 3| < δ → |2x − 1 + 7| < So we want |2x + 6| < , which is the same as 2|x + 3| < , which happens if |x + 3| < /2Claim: δ = /2 worksProof:|x + 3| < δ = /22|x + 3| < |2x + 6| < 3. (4 pts) Prove, using the − δ definition of a limit, that limx→16√x = 4Let > 0 be given. We want to find δ > 0 st 0 < |x − 16| < δ ⇒ |√x − 4| < |√x − 4| = |√x − 4||√x + 4||√x + 4|=|x − 16|√x + 4Note also that if |x −16| < 1, then 15 < x < 17, so√15 + 4 <√x + 4 <√17 + 4. Since the√x + 4 is in the denominator, we’ll replace it by the smaller thing:|x − 16|√x + 4<|x − 16|√15 + 4We want this to be < , so we’ll guess δ = min(1, (√15 + 4))Claim: This worksProof: If |x − 16| < δ = min(1, (√15 + 4)), then |x − 16| < 1 and |x − 16| < (√15 + 4)).Be previous work, we know√x + 4 >√15 + 4 under these conditions, so we can do thefollowing:|√x − 4| = |√x − 4||√x + 4||√x + 4|=|x − 16|√x + 4<|x − 16|√15 + 4<(√15 + 4)√15 + 4= as was to be
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