$f : \mathbb{R} \to \mathbb{R}$ continuous and positive. Show $\lim\limits_{t\to 0^+} \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx =f(0)$ 
$f : \mathbb{R} \to \mathbb{R}$ continuous and positive. Show $$\lim_{t\to 0^+} \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx =f(0)$$

$f$ is continuous so $\forall \epsilon >0\,, \exists \delta > 0, |x| < \delta \implies |f(x) -f(0)| < \epsilon$
And, $f$ is continuous on $[t,t+t^2]$ so $\exists M>0, |f(x)| < M, x \in [t,t+t^2] $ 
As $f$ is positive, I can't wlog take $f(0)=0$.
I've tried two approaches:
1)   $\bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x) - f(0) dx \bigg|$
2)   $\bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x) dx \bigg| = \bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx - \frac{1}{t^3} \int_t^{t+t^2} xf(0)dx + \frac{1}{t^3} \int_t^{t+t^2} xf(0)dx \bigg|$
The first seems to fail because I can't get to $x(f(x)-f(0))$ nicely and the second seems to fail because it doesn't seem to get me to a situation where I have $f(0) + \epsilon$ remaining. 

Hint: $$\frac1{t^3}\int_t^{t+t^2}x\,dx=\frac1{2t^3}((t+t^2)^2-t^2)
=1+\frac t2,$$so $$f(0)-\frac1{t^3}\int_t^{t+t^2}xf(x)\,dx
=-t\frac{f(0)}{2}+\frac1{t^3}\int_{t}^{t+t^2}x(f(0)-f(x))\,dx.$$
