Exchanging limits and Riemann Integral Consider a function $f:\Theta \subseteq \mathbb{R}\rightarrow \mathbb{R}$ continuous at $\theta=\theta_0$ such that $f(\theta)\geq 0$ $\forall \theta \in \Theta$. Consider a sequence of real numbers $\{h_t\}_t$ such that $\lim_{t\rightarrow \infty}h_t=h$. Assume that $f(\theta)$ is Riemann integrable in a neighbourhood of $\theta_0$.

I want to show that $\lim_{t\rightarrow \infty} \int_{0}^1 h_t^2 f(\theta_0+u \frac{ h_t}{t})du=h^2 f(\theta_0)$. 


  My attempt (incomplete):
  
  (i) if such a limit exists, then $\lim_{t\rightarrow \infty} \int_{0}^1
 h_t^2 f(\theta_0+u \frac{ h_t}{t})du=h^2 \lim_{t\rightarrow \infty} \int_{0}^1
 f(\theta_0+u \frac{ h_t}{t})du$
  
  (ii) By continuity of $f(\theta)$ at $\theta_0$,
   $\lim_{\theta\rightarrow \theta_0} f(\theta)=f(\theta_0)$; moreover, 
   $\lim_{\theta\rightarrow \theta_0} f(\theta)=\lim_{t\rightarrow \infty}
 f(\theta_0+u \frac{ h_t}{t})$ since $\lim_{t\rightarrow \infty}\theta_0+u \frac{ h_t}{t}=\theta_0+u*0=\theta_0$; hence, $\lim_{t\rightarrow \infty}
 f(\theta_0+u \frac{ h_t}{t})=f(\theta_0)$ for any $u$
  
  (iii) Is there a way to exchange integrability and limit? The fact that $f(\theta)$ is Riemann integrable in a neighborhood of $\theta_0$ means that $f(\theta_0+u\frac{h_t}{t})$ is bounded when $u\in [0,1]$. Does this help?
Unless I mean missing something, what helps here is the uniform convergence of the sequence $\left(f\left(\theta_0+u  h_t/t\right)\right)_{t\geqslant 1}$ to the constant function equal to $f(\theta_0)$. This is due to the fact that $f$ is continuous at $\theta_0$: for a fixed positive $\varepsilon$, we can find a $\delta$ such that if $|x|\lt \delta$, then 
$\left|f(\theta_0+x)-f(\theta_0)\right|\lt \delta$. For $t$ large enough and uniformly in $u\in [0,1]$, we have $|u  h_t/t|\leqslant |h_t|/t\lt \delta$.
Unless I mean missing something, what helps here is the uniform convergence of the sequence $\left(f\left(\theta_0+u  h_t/t\right)\right)_{t\geqslant 1}$ to the constant function equal to $f(\theta_0)$. This is due to the fact that $f$ is continuous at $\theta_0$: for a fixed positive $\varepsilon$, we can find a $\delta$ such that if $|x|\lt \delta$, then 
$\left|f(\theta_0+x)-f(\theta_0)\right|\lt \delta$. For $t$ large enough and uniformly in $u\in [0,1]$, we have $|u  h_t/t|\leqslant |h_t|/t\lt \delta$.