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#

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#.

Here's how we would rewrite our integral:

#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#

While the first integral can be solved fairly easily, the second integral requires us to substitute in that previously created Maclaurin series.

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

We can integrate quite easily now.

#= 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)]#

We can remove the second set of parentheses since it contains only zeros.

#= 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#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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