Computing limits of integrals of continuous periodic function Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and satisfy $f(x+1) =f(x)$.
Also assume that $\int_{0}^{1}f = 2$ and $f(0) = 3$
I'm wondering how to compute the following limits.

$\lim_{c \to \infty} \int_{0}^{1}f(cx) dx$
$\lim_{c \to 0} \int_{0}^{1}f(cx) dx$,
where $c$ ranges over the reals.

I'm very rusty with calculus and am unsure of how to proceed. Any help is appreciated.

$t=xc$, $dt= cdx$
$$\int_0^1 f(xc)dx = \frac{1}{c}\int_0^cf(t)dt = \frac{1}{c}(\lfloor c\rfloor \int_0^1f(t)dt+\int_{\lfloor c\rfloor}^cf(t)dt) $$
This we can take limit of as $c\to \infty$, and it's $\int_0^1 f(t)dt$.
As $c\to 0$, we see it's just the derivative at $0$ of $F(x)=\int_0^x f(t)dt$, so from the fundamental theorem of calculus, the value is $f(0)=3$
Edit:
$$\int_0^c f(t)dt = \int_0^{\lfloor c\rfloor} f(t)dt+\int_{\lfloor c\rfloor}^cf(t)dt $$
$$\int_0^{\lfloor c\rfloor} f(t)dt = \sum_{k=0}^{\lfloor c\rfloor-1} \int_k^{k+1} f(t)dt = \sum_{k=0}^{\lfloor c\rfloor-1}\int_0^1f(t)dt = \lfloor c\rfloor \int_0^1f(t)dt $$
