the integral becomes :
Another simplier method that i've just thought of :
Suppose the limit exists ,
Perhaps I was wrong , the limit sign is outside the integral
so we cannot put
I am here try to make some corrections ,
By MVT, let
In fact ,we don't need to put
However . after the corrections , I seem to have chosen a terrible detour . Why I thought in this way ...
BTW , just before the bed , I suddenly think of another method , it is also a simple method :
Integration by parts ,
we just need to prove that the remaining integral tends to zero .
Thanks a lot simplependulum and Krizalid for your help!
Also, I realised that the problem can be generalized.
Also, we observe that \forall polynomial P(x):
Now, let f be continuous on [0,1]. Since, the conditions for Weierstrauss Approximation theorem
are satisfied, we can now establish that
Given and for we have some polynomial such that
Taking limits we see
That is is near and consequently near .
Since was arbitrary, we are done.