Question

How to calculate this? `lim_(n-> oo)int_0^1x^n/(x^n+1)dx`

Answer 1

`lim_(n->oo)int_0^1(x^n)/(x^n+1)dx = 0`

Explanation:

First, let's look at the Maclaurin series for `1/(1+x)`:

`1/(1+x) = 1 - x + x^2 - x^3 + x^4 + cdots`

We know this because `1/(1-x)` is the geometric sum (`1+x+x^2+x^3+cdots`), and this is what we get when we plug in `-x` for `x`.

Now, let's plug in `x^n` for `x`:

`1/(1+x^n) = 1 - x^n + x^(2n) - x^(3n) + x^(4n) - cdots`

Note that the interval of convergence for this sum is `-1 < x < 1`, since we know `-1 < x^n < 1`. We know this because in the original series we substituted, the interval was also `-1 < x < 1`. All this means is that we will be able to use this series to help us find our limit, since our integral only goes from `0` to `1`.

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Let's rewrite our integral like this:

`int_0^1(x^n)/(x^n+1)dx = int_0^1(x^n+1-1)/(x^n+1)dx`

`= int_0^1(x^n+1)/(x^n+1)dx - int_0^1 1/(x^n+1)dx`

`= int_0^1dx - int_0^1 1/(x^n+1)dx`

The first integral is pretty easy to solve. For the second integral we'll need to substitute in that that Maclaurin series we made earlier.

`= 1 - int_0^1 (1 - x^n + x^(2n) - x^(3n) + x^(4n) - cdots)dx`

Now we can integrate fairly easily.

`= 1 - [x - x^(n+1)/(n+1) + x^(2n+1)/(2n+1)-x^(3n+1)/(3n+1)+cdots]_0^1`

`= 1 - [(1 - 1/(n+1) + 1/(2n+1) - 1/(3n+1)+cdots) - (0 - 0/(n+1) + 0/(2n+1) - 0/(3n+1)+cdots)]`

Notice the second set of parentheses is all zeros. We can just get rid of it.

`= 1 - [1 - 1/(n+1) + 1/(2n+1) - 1/(3n+1)+cdots]`

`= 1 - 1 + 1/(n+1) - 1/(2n+1) + 1/(3n+1) + cdots`

`= 1/(n+1) - 1/(2n+1) + 1/(3n+1) + cdots`

So this is our evaluated integral, in terms of `n`. Now -- the problem asks for us to find the limit of this integral as `n` approaches `oo`. So, let's do that:

`lim_(n->oo)(1/(n+1) - 1/(2n+1) + 1/(3n+1) + cdots)`

It should be fairly clear that as `n` goes to `oo`, all of the denominators will also go to `oo`, and we know that when the numerator is finite and the denominator is infinite, the whole fraction will just be `0`. So this means that:

`lim_(n->oo)(1/(n+1) - 1/(2n+1) + 1/(3n+1) + cdots)`

`= 0 - 0 + 0 - 0 + 0 - cdots = 0`

So as we approach infinity, the integral approaches `0`.

Final Answer