let f be continous on [0,1] and suppose f(0)=0.
Show that

[lim  (integral(f(x^n)dx)) from 0 to 1] = 0
n goes infinity

Hi fmunshi!
In order to answer this question, it's best to use the limit
definition of an integral. You can find it in the first page of the
following pdf document.

The Definite Integral
http://homepages.ius.edu/WCLANG/m215/notes5-2.pdf

Let's apply that definition to this particular problem. As you can see
in the document, "delta x" is defined as (b-a)/n, where b is the upper
limit of the integral and a is the lower limit (in your case a=0 and
b=1), and 'n' is the number of partitions of the interval [a,b]. Since
in your question you use 'n' for other purposes, let's call 'p' the
number of partitions of the interval.

So, integral(f(x^n)dx) from 0 to 1 is, by definition,

          p
lim      ----
p->inf.  \     f( (xi*)^n ) * (1-0)/p
         /
         ----
          i=1

where xi* is as defined in the document: a sample point in the i-th
subinterval defined by the partition. What you want to calculate then
is actually

                   p
lim      lim      ----
n->inf.  p->inf.  \     f( (xi*)^n ) * 1/p
                  /
                  ----
                   i=1

We can then use a property of the limit to "exchange" them in the
following way

                   p
lim      lim      ----
p->inf.  n->inf.  \     f( (xi*)^n ) * 1/p
                  /
                  ----
                   i=1

Notice that now we can first calculate the limit of the expression as
n goes to infinity, and then take the limit of the result as p goes to
infinity. Using another standard property of the limit, we can put the
lim inside the summatory, to get

           p
lim      ----
p->inf.  \     lim      f( (xi*)^n ) * 1/p
         /     n->inf.
         ----
          i=1

Now, since xi* are in the interval [0,1], then, as n goes to infinity,
(xi*)^n goes to 0. The only exception would occur if we took the x* of
the last sub-interval to be 1. We would get that this x* goes to 1.
But we'll see that this is not an important issue.

We have then that (xi*)^n goes to either 0 or 1 (in the exception
case). Now, since f is continuous, if (xi*^n) goes to 0, then f(xi*^n)
goes to f(0) (or f(1) in the other case). Since f(0)=0 and f(1)=L (any
number), then we get that the summatory becomes either

f(1)*1/p  (in the exception case)
or
0         (if the x* of the last sub-interval is not taken to be 1)

This is because all the terms (except maybe the last one) become f(0),
which is equal to 0. Finally, it's clear that either

lim        f(1)*1/p  = 0
p->inf.
or
lim        0         = 0
p->inf.

Therefore, in any case,

[lim  (integral(f(x^n)dx)) from 0 to 1] = 0


I hope this helps! If you have any doubts regarding this answer,
please request a clarification before rating it.

Best wishes!
elmarto

I think the following is a simpler method of approach:

We show that for any r > 0, there is an N > 0 such that for any n > N,
the absolute value of the above integral is less than r.

[Note: this is the definition of the limit being 0 as n -> infinity.]

Pick any r > 0. Since f is continuous on the closed interval [0,1], it
is bounded and hence has a maximum absolute value; call this M. If M =
0 then f is the zero function on [0, 1] and the limit is obviously 0.
So we can take M > 0 and let k = min{r/(2M), 1/2}.

Since f is continuous, we can find d > 0 such that whenever x is in
[0, d], abs(f(x)) < r/2. [Note: I'm using abs() for the absolute value
function, since the vertical bars as in |x| may be confusing in plain
text format.] Now 1-k is strictly between 0 and 1 (since k > 0 and k
<= 1/2), so (1-k)^n goes to 0 as n goes to infinity. So we can pick a
positive integer N such that (1-k)^N < d.

Then, for any n > N,

abs(integral([f(x^n) dx], x=0..1))
  <= integral([abs(f(x^n)) dx], x=0..1)
  = integral([abs(f(x^n)) dx], x=0..1-k) + integral([abs(f(x^n)) dx],
x=1-k..1)
  <= integral([(r/2) dx], x=0..1-k) + integral([M dx], x=1-k..1)
  = (r/2) (1-k) + Mk
  = (r/2) - (kr/2) + (r/2)
  = r - kr/2 
  < r.

Hence the limit is 0.

----
A comment about my comment:

The essence of this approach is that we can manage the contribution
near x = 0 because f is continuous and f(x) = 0; we can squash
arbitrarily large portions of the interval [0, 1] close to 0 by
raising to a sufficiently large power; and since f is bounded this
allows us to manage the contribution near x = 1 by making that part of
the interval small. If f was continuous only on [0,1) we couldn't do
this, since it would be possible for the value of f(x) as x->1 to rise
faster than we could squash x down to 0. It's probably a good exercise
to find a function which exhibits this behaviour...

-- 
The Scarlet Manuka

I think there is a problem with the "official" google answer.  Elmato
claims that the following reasoning is OK:

"
 
                   p 
lim      lim      ---- 
n->inf.  p->inf.  \     f( (xi*)^n ) * 1/p 
                  / 
                  ---- 
                   i=1 
 
We can then use a property of the limit to "exchange" them in the
following way
 
                   p 
lim      lim      ---- 
p->inf.  n->inf.  \     f( (xi*)^n ) * 1/p 
                  / 
                  ---- 
                   i=1 
 
"

Elmato seems to be saying that you can commute the Limit operators. 
That is not true in general.  For example,

lim      lim      p/(n+p)   = 0
p->inf.  n->inf.  

does not equal

lim      lim      p/(n+p)   = 1.
n->inf.  p->inf.