I have that $ f(x,t) = \frac{2x^2t}{(x^2 + t^2)^2}, t \in [0,1] $ and I am asked to show that $$ \lim_{x \downarrow 0} \int_0^1 f(x,t) \,dt \neq \int_0^1 \lim_{x \downarrow 0} f(x,t) \,dt $$
I have that the RHS is 1 and the LHS is 0.
I am struggling to answer is why this doesn't work in terms of interchanging the limit and the integral. I think it may be to do with the fact $f(x,t)$ doesn't converge uniformly to $f(x)$ but I don't know how to show this or if it is even the correct reason why the interchanged limit and integral aren't equal.
How would I prove $f_t(x)$ doesn't converge to $f(x)$ uniformly (that is assuming I have correctly identified why it doesn't work)?
Set $x=1/n$. Then you have $\displaystyle\int_0^1\frac{2(1/n)^2t}{( (1/n)^2 + t^2)^2}dt$ and the fact that the two integrals are not equal proves that the convergence of the integrand sequence is not uniform.
To do it from scratch, note that $f_n(t)=\frac{2(1/n)^2t}{( (1/n)^2 + t^2)^2}\to 0$ pointwise but with $t_n=1/n,$ we get $\frac{2/n^3}{2/n^4}=n\to \infty$ as $n\to \infty$, so the convergence is not uniform.