Let $\lim_{t\to\infty}b(t) = 0$. Show: All solutions of $\dot{x}(t) + x(t) = b(t)$ converge to $0$. Assignment:

Let $b: \mathbb{R} \rightarrow \mathbb{R}$ be continuous with $\lim_{t\to\infty}b(t) = 0$. Show that all solutions for the ODE $$\dot{x}(t) + x(t) = b(t)$$
  converge for $t\to\infty$ to $0$.

What I have is that $$x(t) = \int_{t_0}^{t} b(\tau)e^{-(t-\tau)}d\tau + c\cdot e^{-t+t_0} = e^{-t} \int_{t_0}^{t} b(\tau)e^{\tau} + c\cdot e^{-t+t_0}$$
where the second part goes to $0$ for $t$ to infinity. However, how can I show that the first part does so as well? I've tried to use different inequalities and identities (also the MVT) but I cannot get it to work.
I'd appreciate any help.
 A: Note that
$| \int_{t_1}^{t_2} b(s) e^{-(t_2-s)} ds | \le \max_{t \in [t_1,t_2]}|b(t)| \le \sup_{t \in [t_1,\infty)}|b(t)|$.
Let $\epsilon>0$. Since $b(t) \to 0$, we can pick some $t'$ such that $| \int_{t_1}^{t_2} b(s) e^{-(t_2-s)} ds | < {1 \over 2} \epsilon$ as long
as $t_1 \ge t'$.
Suppose $t \ge t'$ and write $\int_{t_0}^{t} b(s) e^{-(t-s)} ds = \int_{t_0}^{t'} b(s) e^{-(t-s)} ds + \int_{t'}^{t} b(s) e^{-(t-s)} ds $.
We see that $b(s) e^{s}$ is bounded on $[t_0,t']$, and 
$\int_{t_0}^{t'} b(s) e^{-(t-s)} ds = e^{-t}\int_{t_0}^{t'} b(s) e^{s} ds$, hence we can choose $t^{''} \ge t'$ such that 
$\int_{t_0}^{t'} b(s) e^{-(t-s)} ds < {1 \over 2} \epsilon$ whenever $t \ge t^{''}$.
Hence 
$\int_{t_0}^{t} b(s) e^{-(t-s)} ds \to 0$.
A: Since $b \to 0$, you can find $T > 0$ such that $b(t) < \epsilon$ for all $t > T$. 
Then 
$$ \int_0^t b(\tau) e^\tau \mathrm{d}\tau = \int_0^T b(\tau) e^\tau \mathrm{d}\tau + \int_T^t b(\tau) e^\tau \mathrm{d}\tau $$
The first term is a constant which depends on $T$, the second term can be bounded by 
$$ \int_T^t \epsilon e^\tau \mathrm{d}\tau = \epsilon e^{t - T} $$
Therefore you can find a $T'$ sufficiently large (much larger than $T$) such that for all $t > T'$
$$ |e^{-t} \int_0^t b(\tau) e^\tau \mathrm{d}\tau| \leq C_T e^{-T'} e^{-t + T'} + \epsilon e^{-T} < \epsilon$$ 
