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MAT136H5 S 2025 - All Sections 5.1 preparation check

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Question: Does the sequence{an}  with an=nsin( 1 n )  converge or diverge? If it converges, find it's limit.  A student hands in the following solution. Is it correct?  Line 1: One way to determine if a sequence converges is to try to evaluate the limit as n⟶∞ . Line 2: lim n→∞an=lim n→∞nsin( 1 n )=  Line 3: lim n→∞ sin( 1 n ) 1 n =  Line 4: Both numerator and denominator approach 0 as n⟶∞ , so we can use l'Hopital's rule: Line 5: lim n→∞ cos(n)(− 1 n2 ) − 1 n2 =  Line 6: lim n→∞cos(n)=∞   Line 7: Because the limit is ∞, this means the sequence diverges. Is the student's solution correct? If not, in which line is the first mistake?  Incorrect, with first error in line 5 If not, what is the correct final answer to the question? Final answer: The sequence converges to 1

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The prompt presents a sequence a_n = n sin(1/n) and asks about convergence, plus a student solution with two listed conclusions. We will evaluate each provided option carefully and point out the reasoning issues in the student’s steps before confirming the true behavior of the sequence. Option 1: "Incorrect, with first error in line 5" - The claim here is that the first incorrect line occurs at line 5. To judge this, we examine the student’s steps up to line 5. The student rewrites the limit as lim_{n→∞} [n sin(1/n)]. They then consider lim sin(1/n)/(1/n), which is a valid reformulation since a_n = n * sin(1/n) = sin(1/n) / (1/n). The observation that both numerator sin(1/n) and den......Login to view full explanation

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