A question on limit with definite integral Could any one help me for this one ?

If $f$ is continuous on $[0,1]$ and $f(0)=1$, then $$\lim\limits_{a\to 0}G(a)=\frac{1}{a}\int_{0}^{a}f(x)dx=?$$

 A: $\frac1a\int_0^af(x)\,\mathrm{d}x$ is the average of $f$ over $[0,a]$. Since $f$ is continuous, as $a\to0$, $f$ on $[0,a]$ is close to $f(0)$. Thus, a good guess would be that $\frac1a\int_0^af(x)\,\mathrm{d}x=f(0)$. Let's add some rigor.
Since $f$ is continuous at $0$, for any $\epsilon>0$, there is a $\delta>0$ so that for all $|x-0|<\delta$, we have $|f(x)-f(0)|<\epsilon$, and then
$$
\begin{align}
\left|\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x-f(0)\right|
&=\lim_{a\to0}\frac1a\left|\int_0^a(f(x)-f(0))\,\mathrm{d}x\right|\\
&\le\lim_{a\to0}\frac1a\int_0^a|f(x)-f(0)|\,\mathrm{d}x\\
&\le\lim_{a\to0}\frac1a\int_0^a\epsilon\,\mathrm{d}x\\
&=\epsilon
\end{align}
$$
Since $\epsilon$ is arbitrary,
$$
\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x-f(0)=0
$$
Therefore,
$$
\lim_{a\to0}\frac1a\int_0^af(x)\,\mathrm{d}x=f(0)
$$
A: Apply L'Hospital: with $G$ a primitive of $f$ in the unit interval,$$\lim_{a\to 0}\frac{1}{a}\int_0^a\,f(x)\,dx=\lim_{a\to 0}\frac{G(a)-G(0)}{a}=\lim_{a\to 0}G'(a)=\lim_{a\to 0}f(a)=1$$ by continuity
