Let $f$ be a continuous function and $f:[0,1]\to\mathbb R$. Find $$\lim_{x\to0}\int_0^1\frac{xf(t)}{x^2+t^2}\,dt$$
I am finding that the limit does not exist. But the question is stated in a way to enable one to think that the limit actually exists. But I don't think the limit exists, the reason being simple:
By Mean Value Theorem we can write the integral as $$xf(c)\int_0^1\frac{dt}{x^2+t^2}=f(c)\tan^{-1}\left(\frac{1}{x}\right)$$
Consider a positive subsequence of $x_n\to0$ for which $f(c)\tan^{-1}(\frac{1}{x})\to f(c)\frac{\pi}{2}$ and for a negative subsequence of $x_n\to0$ we will have $f(c)\tan^{-1}(\frac{1}{x})\to-f(c)\frac{\pi}{2}$. So the limit does not exist.
So is it a situation of mis-statement of a question or am I doing something wrong?
EDIT: I forgot to mention that in my solution, $c\in(0,1)$
Assume for the moment that $x>0$. Then the change of variable $t=xu$ turns the integral into $$ \int_{[0,1/x]} \frac{f(xu)}{1+u^2}du. $$ If $$d\mu = \frac{dx}{1+u^2},$$ then by the dominated convergence theorem with respect to the finite measure (so that constants are integrable!) $\mu$ shows that the last integral converges to $f(0)\mu (\mathbb{R}^+)$.
If $x \in \mathbb{R}$, the constant function $f =1$ shows that the limit does not exist.