How to prove this limit is equal to $\beta/a$? I have to solve the following: 

Let $a\in\mathbb{C}$ such that it's real part is $>0$, and $b$ a continuous function such that $\lim_{x\to\infty}b(x)=\beta$ then $$\lim_{x\to\infty}\frac{\int_{0}^{x}e^{at}b(t)dt}{e^{ax}}=\frac{\beta}{a}.$$

I know how to solve the case $\beta\neq 0$ (using L'hopital rule). But I'm stuck in the case $\beta =0$. 
Any help will be apprecieted. 
 A: Let $q = Re(a)>0$. Pick any $\epsilon > 0$.
Let $\delta > 0$ (the exact value of which we'll determine later). Since $\lim_{x \to \infty} b(x) = 0$, for large enough $u > 0$ we have $|b(x)| < \delta$ for all $x > u$. Thus, 
$$
\left|\int_u^x e^{at} b(t) dt\right| \leq \int_u^x |e^{at} b(t)| dt \leq\int_u^x |e^{at} \cdot \delta| = \leq\int_u^x e^{qt} \cdot \delta = \delta\cdot\frac{e^{qx} - e^{qu}}{q}
$$
therefore
$$
\left|\frac{\int_u^x e^{at} b(t) dt}{e^{ax}}\right| \leq \delta\cdot\frac{e^{qx} - e^{qu}}{qe^{qx}}
$$
Now
$$
\lim_{x \to \infty}\delta\cdot\frac{e^{qx} - e^{qu}}{qe^{qx}} = \frac{\delta}{q}
$$
Let $\delta = \epsilon q$. We have:
$$
\lim_{x \to \infty} \left|\frac{\int_0^x e^{at} b(t) dt}{e^{ax}}\right| \leq \lim_{x \to \infty} \left|\frac{\int_0^u e^{at} b(t) dt}{e^{ax}}\right| + \lim_{x \to \infty} \left|\frac{\int_u^x e^{at} b(t) dt}{e^{ax}}\right| \leq \lim_{x \to \infty} \left|\frac{\int_0^u e^{at} b(t) dt}{e^{ax}}\right| + \lim_{x\to\infty} \delta\cdot\frac{e^{qx} - e^{qu}}{qe^{qx}} = 0 + \epsilon = \epsilon
$$
Since $\epsilon > 0 $ was arbitrary, $\lim_{x \to \infty} \left|\frac{\int_0^x e^{at} b(t) dt}{e^{ax}}\right| = 0$
