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.
 A: 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 $|h|<\eta$,
$$\left|\int_0^1 \frac{f(hx)}{x^2+1}\,dx-f(0)\frac\pi4\right|<\epsilon$$
A: One way to knock this out is the dominated convergence theorem. Take any sequence $(h_n)$ of non-zero numbers such that $h_n \to 0$ and define $$g_n(x) = \frac{f(h_nx)}{1+x^2}, \,\,\,\, x \in [0,1].$$ Then $$g_n(x) \to g(x) = \frac{f(0)}{1+x^2}$$ pointwise on $[0,1]$ and since $f$ is continuous, it is bounded so $$\lvert g_n(x) \rvert \le \frac{M}{1+x^2} \le M$$ for some constant $M$. Since the range of integration is compact, we can apply the dominated convergence theorem to show that $$\lim_{n\to\infty} \int^1_0 g_n(x) dx = \int^1_0 \lim_{n\to \infty} g_n(x) dx = f(0) \frac \pi 4$$ as desired. 
Alternatively, if you need to do it directly, notice that \begin{align*}\left \lvert \int^1_0 \frac{f(hx)}{1+x^2}dx - \int^1_0\frac{f(0)}{1+x^2}dx \right \rvert &= \left \lvert\int^1_0 \frac{f(hx) - f(0)}{1+x^2} dx \right \rvert \\
&\le \int^1_0 \frac{\lvert f(hx) - f(0) \rvert}{1+x^2}dx. \end{align*} Since $f$ is continuous on a compact set, it is uniformly continuous. Hence for any $\epsilon > 0$, there is $\delta > 0$ such that for all $x,y$ with $\lvert x - y \rvert < \delta$, we have $\lvert f(x) - f(y) \rvert < \epsilon$. Then for $0 < h < \delta$, we will also have $0 < hx < \delta$ (for any $x \in (0,1)$) and so $\lvert f(hx) - f(0) \lvert < \epsilon.$ For such $h$, we see \begin{align*}\left \lvert \int^1_0 \frac{f(hx)}{1+x^2}dx - \int^1_0\frac{f(0)}{1+x^2}dx \right \rvert
&\le \int^1_0 \frac{\epsilon}{1+x^2}dx = \epsilon\frac \pi 4. \end{align*} which is the same thing you wanted up to rescaling $\epsilon$.
