Consider the following limit: $$\lim_{t\to 0}\frac{x \sin(xt)}{1+x^2}.$$The question is to determine whether it exists, and if it does, whether it is uniform, i.e. whether there is $\delta_\epsilon>0$ not depending on $x$ such that $\left|\frac{x\sin(xt)}{1+x^2} - L\right|<\epsilon$ for all $0<t<\delta$, for all $x\in S$, for some set $S$ (say $S=\mathbb{R}$).
I think it is clear that the limit is zero for all $x$. In looking at the uniformity, certainly it converges uniformly on a bounded set $\{|x|<B\}$. I conjecture that it doesn't converge uniformly on $x\in \mathbb{R}$. To show that it doesn't, I'm trying to make convenient choices for $t$ to show that as $x$ gets large, a particular nonzero value of $\frac{x \sin(xt)}{1+x^2}$ occur arbitrarily close to $t=0$. For instance, I might choose $t = 1/x$, which gives $\frac{x \sin(1)}{1+x^2}$. But this goes to zero as $x\to \infty$, so this doesn't help.
I notice that $\sin(xt)=xt + O(x^3t^3)$, so $$\frac{x\sin(xt)}{1+x^2}=\frac{x^2 t}{1+x^2}+\frac{O(x^4 t^3)}{1+x^2},$$ which doesn't seem to shed any light.
Any ideas?
Note that $$ \left\lvert\frac{x}{1+x^2}\right\rvert\rightarrow0\text{ as }\lvert x\rvert\rightarrow\infty. $$ So, for any $\epsilon>0$, you can find $X>0$ such that $$ \left\lvert\frac{x\sin(tx)}{1+x^2}\right\rvert\leq\left\lvert\frac{x}{1+x^2}\right\rvert<\epsilon\text{ whenever }\lvert x\rvert\geq X, $$ regardless of the value of $t$.
So, if you can show that convergence is uniform on $\{\lvert x\rvert\leq X\}$, you are golden!