Limit of integrals of trigonometric functions

Question

If we have a sequence of functions defined by $f_n(x) = \frac{\sin((x + \frac 1n)^2) - \sin((x-\frac 1n)^2}{\sin(\frac 1n)},$ how can we find $\lim_{n \rightarrow \infty} \int_0^1 f_n(x) \, d\lambda$ where $\lambda$ is the lebesgue measure? I think that the limit of the $f_n(x)$ is twice the derivative of $\sin(x^2)$ so $f_n(x) \rightarrow 4x\cos(x^2)$ but I don't know what dominating function we would use to apply the dominated convergence theorem. It also looks like it might be an application of the Lebesgue differentiation theorem, but I'm completely lost on how to apply it. Am I on the right track, or do I need to look at something completely different?

Answer 1

Hint: Suppose $f$ is continuously differentiable on $[-1,2].$ Then for any $x\in [0,1]$ and $h\in (0,1],$

$$\tag 1 \frac{f(x+h)-f(x-h)}{2h} = f'(c(x,h)),$$

by the MVT. Here $c(x,h)$ lies between $x-h$ and $x+h.$ Thus a dominating function for the left side of $(1)$ is just the constant $\sup_{[-1,2]}|f'|.$