Find the limit

i simply expanded e^(stuff) as a power series. all terms except the 1st can be ignored. So you get 1/3 :D

We have

x^{2}>x^{2}e^{-\frac{x^{2}}{n^{2}}}>x^{2}e^{-\frac{1}{n^{2}}}

Integrating and taking limit we get the result as 1/3 by sandwich theorem..

okie..e^(stuff) = 1 + (stuff) + (stuff)^2/2! + (stuff)^3/3! .....he he

Do you think this is correct..

given stuff = \int_{0}^{1}{}\lim_{n\rightarrow \infty}x^{2}e^{-\frac{x^{2}}{n^{2}}}dx = \int_{0}^{1}{}x^{2}dx = \frac{1}{3} ??

that is why i humbly said simple-minded guess. I have seen it being used but i am not aware under what conditions it is valid. I know something called Fubini's theorem comes into play, but details are hazy

The fact that the integral converges is enough to guarantee the operation of bringing the limit inside the integral .

The actual concept of what nasiko is saying goes something like this :

If {f_n} denote a sequence of functions converging uniformly to a function f then the following can be said -

\lim_{n\rightarrow \infty}\sum_{x=a}^{b}{f_n(x)}=\sum_{x=a}^{b}{\lim_{n\rightarrow \infty}f_{n}(x)}

The concept of integration can be said equivalently.

The uniform convergence concept is somewhat like this, Suppose x\in [a,b].

If there exists a positive integer M, such that for all n>M, and for all x in [a,b]

|f_n(x)-f(x)|\le \epsilon

where ε is an arbitary positive quantity then we can cocnclude that
\left\{f_n(x) \right\}\rightarrow f(x) uniformly.

There are many ways to test uniform convergence of functions, I can put down some if you need any.

Edit: There are two types of convergence of sequences of functions - piecewise and uniform. Piecewise convergence is never sufficient for the operation stated above.