$f : \mathbb{R} \to \mathbb{R}$ 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. 
 A: Denote $\alpha_{t}$ and $\beta_{t}$ be such that $f(\alpha_{t})=\min_{x\in[t,t+t^{2}]}f(x)$, and $f(\beta_{t})=\max_{x\in[t,t+t^{2}]}f(x)$, then $xf(\alpha_{t})\leq xf(x)\leq xf(\beta_{t})$ on $[t,t+t^{2}]$. Taking integral each side with quotient to $t^{3}$, one finds 
\begin{align*}
\dfrac{1}{t^{3}}\int_{t}^{t+t^{2}}xdx=\dfrac{1}{2}t+1\rightarrow 1
\end{align*}
as $t\rightarrow 0^{+}$. Now $f(\alpha_{t}),f(\beta_{t})\rightarrow f(0)$ as $t\rightarrow 0^{+}$.
A: Observes that 
$$ \frac{1}{t^3} \int_t^{t+t^2} xf(0) = (1+\frac{t}2)f(0)$$ 
Assume 
$$|f(x)-f(0)|\le \varepsilon~~~~for ~~~~0<x<\delta$$
Then take $0<t<\min(\varepsilon, \frac{\delta}{1+\varepsilon})~~~$ then $0<t<t+t^2\le t(1+\varepsilon)<\delta$ thefore,
$$\color{blue}{|f(x)-f(0)|\le \varepsilon~~~~for ~~~~ t<x<t+t^2<\delta}$$ Whence,
$$\bigg |\frac{1}{t^3} \int_t^{t+t^2} xf(x)dx-f(0) \bigg|\\= \bigg |\frac{1}{t^3} \int_t^{t+t^2} x\bigg(f(x)-f(0)\bigg) dx+\frac{1}{t^3} \int_t^{t+t^2} xf(0)dx -f(0) \bigg|\\=\bigg |\frac{1}{t^3} \int_t^{t+t^2} x\bigg(f(x)-f(0)\bigg) dx+\frac{t}{2} f(0) \bigg|
\\\le \frac{1}{t^3} \int_t^{t+t^2} x|f(x)-f(0)| dx + \bigg|\frac{t}{2} f(0)  \bigg|< \varepsilon (1+\frac{t}2)+\bigg|\frac{t}{2} f(0)  \bigg| <\varepsilon (1+\frac{\varepsilon}2)+\bigg|\frac{\varepsilon}{2} f(0)  \bigg| \to0$$
A: Hint: $$\frac1{t^3}\int_t^{t+t^2}x\,dx=\frac1{2t^3}((t+t^2)^2-t^2)
=1+\frac t2,$$so $$f(0)-\frac1{t^3}\int_t^{t+t^2}xf(x)\,dx
=-t\frac{f(0)}{2}+\frac1{t^3}\int_{t}^{t+t^2}x(f(0)-f(x))\,dx.$$
A: 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.
