Limit of $\int_0^1\frac{f(hx)}{x^2+1}dx$ when $h\to0$ 
Let $f\in \mathcal{C}^0\big([0,1],\mathbb{R}\big)$ and, for every $h\in(0,1]$, $$I(h)=\int_0^1\dfrac{f(hx)}{x^2+1}dx$$
For $\varepsilon >0$, show there exists $\eta>0$ such that for every $h\in(0,\eta)$,
$$\left|I(h)-f(0)\frac{\pi}{4}\right|\leq \varepsilon$$

Since $\dfrac{\pi}{4} = \displaystyle\int_0^1\dfrac{1}{x^2+1}dx$
$$\left |\int_0^1\dfrac{f(hx)}{x^2+1}dx -\int_0^1\dfrac{f(0)}{x^2+1}dx \right|\leq\varepsilon$$
And now, I have got no idea how to solve this problem. I think, I should show that :
$$|h|\leq \eta \implies \left|I(h)-f(0)\frac{\pi}{4}\right|\leq \varepsilon$$
but I'm not sure.

Note that we can write
$$\begin{align}
\left|\int_0^1 \frac{f(hx)}{x^2+1}\,dx-f(0)\frac\pi4\right|&=\left|\int_0^1 \frac{f(hx)-f(0)}{x^2+1}\,dx\right|\\\\
&\le \int_0^1 \frac{|f(hx)-f(0)|}{x^2+1}\,dx\\\\
&
\le\frac\pi4 \sup_{x\in [0,1]}|f(hx)-f(0)|
\end{align}$$
For any $\epsilon>0$, there exists a number $\eta>0$ such that $|f(hx)-f(0)|<4\epsilon/\pi$ whenever $|hx|<\eta$.  
Since $x\in [0,1]$, then whenever $|hx|<|h|<\eta$, we have $|f(hx)-f(0)|<4\epsilon/\pi$.  
Hence, $\sup_{x\in [0,1]}|f(hx)-f(0)|<4\epsilon/\pi$ whenever $|h|<\eta$.
Putting it all together, we have for $|h|<\eta$,
$$\left|\int_0^1 \frac{f(hx)}{x^2+1}\,dx-f(0)\frac\pi4\right|<\epsilon$$
