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Problem 7.2.5. Prove Theorem 7.2.4. Theorem 7.2.4. If lim an = a and lim b, = b, then lim (a, + b,) = a + b.

Problem 7.2.5. Prove Theorem 7.2.4. Theorem 7.2.4. If lim an = a and lim b, = b, then lim (a, + b,) = a + b.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Real math analysis. Problem 7.2.5 please 

10:48 SCRAPWORK: Given ɛ > 0, we want N so that if n > N, then a,bn the standard tricks in analysis is to ab < e. One of "uncancel." In this case we will subtract and add a convenient term. Normally these would "cancel out," which is why we say that we will uncancel to put them back in. You already saw an example of this in proving the Reverse Triangle Inequality 7.2.3). In the present case, consider |a,b, – ab| = |a,b, – a„b+ a„b – ab| < lanb, – amb| + |a„b – ab| = |an||bn – 6| + |||an – al. We can make this whole thing less than e, provided we make each term in the sum less than 5. We can make b an – a| < if we make an – a < 2. But wait! What if b = 0? We could handle this as a separate case or we can do the following "slick trick." Notice that we can add one more line to the above string of inequalities: |an||bn – b| + |b||a, – al < |an||bn – b| + (|b| + Now we can make |an – a| < 2(6+1) and not worry about dividing by zero. bit II
Transcribed Image Text:10:48 SCRAPWORK: Given ɛ > 0, we want N so that if n > N, then a,bn the standard tricks in analysis is to ab < e. One of "uncancel." In this case we will subtract and add a convenient term. Normally these would "cancel out," which is why we say that we will uncancel to put them back in. You already saw an example of this in proving the Reverse Triangle Inequality 7.2.3). In the present case, consider |a,b, – ab| = |a,b, – a„b+ a„b – ab| < lanb, – amb| + |a„b – ab| = |an||bn – 6| + |||an – al. We can make this whole thing less than e, provided we make each term in the sum less than 5. We can make b an – a| < if we make an – a < 2. But wait! What if b = 0? We could handle this as a separate case or we can do the following "slick trick." Notice that we can add one more line to the above string of inequalities: |an||bn – b| + |b||a, – al < |an||bn – b| + (|b| + Now we can make |an – a| < 2(6+1) and not worry about dividing by zero. bit II
10:44 Theorem 7.2.4. If lim an = a and lim b, = b, then lim (a, + b„) = a + b. We will often informally state this theorem as "the limit of a sum is the sum of the limits." However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We'll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof. Problem 7.2.5. Prove Theorem 7.2.4. Theorem 7.2.4. If lim an = a and lim b, = b, then lim (an + bn) = a + b. in-context SCRAPWORK: If we let ɛ > 0, then we want N so that if II II
Transcribed Image Text:10:44 Theorem 7.2.4. If lim an = a and lim b, = b, then lim (a, + b„) = a + b. We will often informally state this theorem as "the limit of a sum is the sum of the limits." However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We'll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof. Problem 7.2.5. Prove Theorem 7.2.4. Theorem 7.2.4. If lim an = a and lim b, = b, then lim (an + bn) = a + b. in-context SCRAPWORK: If we let ɛ > 0, then we want N so that if II II
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