2018-12-15T23:28:03ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=24701In fact the "exact" solution can be obtained by the following argument:
since 0<x<1, but we have:
which is the exact value of the limit. (sorry for the confusion)
]]>http://www.mathisfunforum.com/profile.php?id=2122842018-12-15T23:28:03Zhttp://www.mathisfunforum.com/viewtopic.php?pid=406451#p406451In the previous solution, we have taken the average of the denominator as the weight in the summation which provides a good approximation of the limit, but it does not produce the exact solution. To get a more exact solution we should take as the weight in the summation the average of the whole fraction in the integrand. This follows from the following argument:
but we have:
Therefore, we get
and hence ]]>http://www.mathisfunforum.com/profile.php?id=2122842018-12-15T18:05:51Zhttp://www.mathisfunforum.com/viewtopic.php?pid=406449#p406449Since 0<x<1 we notice that:
we would have
Therefore we have:
so the limit is between 1/6 and 1/2. However, if we calculate the average value of the function in the denominator, we would get:
but the average, by definition, is the amount that produces the same sum if each term in the summation is replaced by it, therefore we have:
and hence
which is the required limit. Notice that it is between 1/6 and 1/2 as predicted. I hope that answers your question.