$f : \mathbb{R} \to \mathbb{R}$ 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. 

Here is a simpler and direct approach. 

We use the generalized mean value theorem for integrals to get $$\int_{t}^{t+t^2}xf(x)\,dx=f(c)\int_{t}^{t+t^2}x\,dx=f(c)(t^3+t^4/2)$$ for some $c\in[t, t+t^2]$ and then letting $t\to 0^{+}$ we see that $c\to 0$ and thus by continuity we have the desired limit as $f(0)$. Note that we only need continuity of $f$ and not necessarily $f$ to be positive.
