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**RaduVF****Guest**

Let

Calculate

**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

Since 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.

Best wishes

*Last edited by Grantingriver (2018-12-15 04:06:22)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

In 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

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

In 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)

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