Limit of Sequence: Find n^2(e^\frac{1}{n^2} - cos(\frac{1}{n}))

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The limit of the expression n^2(e^(1/n^2) - cos(1/n)) is discussed, with an initial incorrect assumption leading to a result of 1 instead of the correct limit of 3/2. The mistake arose from misapplying asymptotic behavior, where it was assumed that cos(1/n) could be substituted directly with 1. A suggestion to change variables to x = 1/n and expand both terms in a series around zero clarified the problem. The correct approach involved recognizing that e^(1/n^2) - cos(1/n) behaves differently than initially thought. Ultimately, careful series expansion reveals the accurate limit.
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Homework Statement



Find the limit of n^2(e^\frac{1}{n^2} - cos(\frac{1}{n}))

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The Attempt at a Solution



since cos(1/n) is asymptotic to 1. n^2(e^\frac{1}{n^2} - cos(\frac{1}{n})) ~ n^2(e^\frac{1}{n^2} - 1) ~ n^2 \frac{1}{n^2}) = 1
The right answer is 3/2 though. I don't see what's wrong with my reasoning. Maybe i used asymptotic in an illegitimate way. What's the problem?
 
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You have to be a little more careful than that.
Try switching over to x = 1/n, then it will be the limit for x going to zero.
If you expand both terms inside the brackets in a series around 0, you can throw away terms of order x4 and you will arrive at the right answer.
 
Thanks, that way i solved it.
I also found what i did wrong with asymptotic. I though that when a sequence is asymptotic with another you could just substitute one with the other. But it's not true. in this case. cos(1/n)~ 1 but e^\frac{1}{n^2} - cos(\frac{1}{n}) ~\frac{3}{2} e^\frac{1}{n^2} - 1.

e^\frac{1}{n^2} - cos(\frac{1}{n}) ~e^\frac{1}{n^2} - 1 This is not true.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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