Question: Is $C\_0(\mathbb{R})\subseteq L^1(\mathbb{R})$?

It seems as though it might be because the integral will tend to $0$ at some point, but I cannot prove it.

I thought of splitting as such:

$$ \int\_{-\infty}^\infty |f(x)| dx = \int\_{-M}^M |f(x)| dx + \int\_{M}^\infty |f(x)| dx+ \int\_{-\infty}^{-M} |f(x)| dx$$

but if I bound the last two with some $\epsilon>0$ it's still too large a bound.

Answer: 
Consider the function $F : \mathbb{R} \to \mathbb{R}$ given by $F(0) = 1$ and $F(x) = \frac{\sin x}{x}$ for all $x\neq 0$. Then $F\in C\_0(\mathbb{R})$, but $\int\_{-\infty}^\infty |F(x)|\, dx = \infty$. Indeed, if $n$ is a positive integer, $$\int\_{(n-1)\pi}^{n\pi} |F(x)|\, dx = \int\_0^\pi \frac{\sin x}{x + (n-1)\pi}\, dx \ge \int\_0^\pi \frac{\sin x}{\pi + (n-1)\pi} = \frac{2}{n\pi}$$ and $\sum\_{n = 1}^\infty \frac{2}{n\pi} = \infty$.
Vanishing at infinity is not sufficient for integrability. If (for example) $f(x) \sim 1/x$ for $x \to \infty$ then $f$ is not integrable.

A simple concrete counterexample is
$$
 f(x) = \frac{1}{1+|x|}
$$
which is continuous with $\lim\_{x \to \pm \infty} f(x) = 0$. But
$$
 \int\_0^x |f(t)| \, dt = \ln(1+x) \to \infty
$$
for $x \to +\infty$ shows that $f \notin L^1(\Bbb R)$.