Prove that the solution of $y'+y=\arctan(e^x), y(0)=2$ admits horizontal asymptote. Let us consider the Cauchy problem: 
$$y'+y=\arctan(e^x),\ \ \ \ y(0)=2$$ 
Prove that the function $y(x)$ admits horizontal asymptote without solving the problem.
 A: I'll prove a more general case, instead. If $\lim_{x\rightarrow+\infty}f(x)+f'(x)=l$, then:
$$\begin{align}
&\lim_{x\rightarrow+\infty}f(x)=l\\
&\lim_{x\rightarrow+\infty}f'(x)=0
\end{align}$$
Suppose we have $$\lim_{x\rightarrow+\infty}f(x)+f'(x)=l$$
Then we have $$e^x(f(x)+f'(x))\approx_{x\rightarrow +\infty} le^x $$
$$(e^xf(x))'\approx_{x\rightarrow +\infty} le^x=(le^x)'$$
Since $x\rightarrow le^x$ is positive and its integral diverges at $+\infty$, then
$$e^xf(x)\approx_{x\rightarrow +\infty} le^x \tag{justification at the bottom}$$
Therefore $$f(x)\approx_{x\rightarrow +\infty} l $$
So, finally we have:
$$\begin{align}
&\lim_{x\rightarrow+\infty}f(x)=l\\
&\lim_{x\rightarrow+\infty}f'(x)=0
\end{align}$$

In your example, we have $\lim_{x\rightarrow+\infty} y+y'=\pi/2$
Therefore
$$\begin{align}
&\lim_{x\rightarrow+\infty}y=\pi/2\\
&\lim_{x\rightarrow+\infty}y'=0 
\end{align}$$
and $y$ has an horizontal asymptote.

Justification:
Suppose we have two functions $f,g\in \mathcal C^1$ such that $g'$ is non-negative, and $g(x)$ tends to $+\infty$ as $x$ tends to $+\infty$ and $$f'(x)\approx_{x\rightarrow +\infty} g'(x) $$
i.e $$\forall \varepsilon>0, \exists x_0\in \Bbb R \text{ s.t } x\geq x_0\implies |f'(x)-g'(x)|\leq \varepsilon g'(x) $$
Let $\varepsilon>0$. Then $\exists x_0\in \Bbb R \text{ s.t } x\geq x_0\implies |f'(x)-g'(x)|\leq \frac{\varepsilon}{2} g'(x)$
We have for $x\geq x_0$
$$\begin{align}
|f(x)-g(x)|&=\left|f(0)-g(0)+\int_{0}^x (f'(t)-g'(t))dt\right| \\
&=\left|f(0)-g(0)+\int_0^{x_0} (f'(t)-g'(t))dt + \int_{x_0}^x (f'(t)-g'(t))dt\right|\\
&\leq \left|f(0)-g(0)+\int_0^{x_0} (f'(t)-g'(t))dt\right| + \frac{\varepsilon}{2}\int_{x_0}^x g'(t)dt\\
&\leq \varepsilon g(x)\left(\frac{1}{2}+\frac{\left|f(0)-g(0)+\int_0^{x_0} (f'(t)-g'(t))dt\right|-\frac{\varepsilon}{2}g(x_0)}{\varepsilon g(x)}\right)
\end{align}$$
Since $\frac{1}{g(x)}\rightarrow 0$, then $\exists x_1\in \Bbb R: x\geq x_1\implies \left|\frac{\left|f(0)-g(0)+\int_0^{x_0} (f'(t)-g'(t))dt\right|-\frac{\varepsilon}{2}g(x_0)}{\varepsilon}\right|\frac{1}{g(x)}\leq \frac12$
So for $x\geq\max(x_0,x_1)=x_2$, we have:
$$|f(x)-g(x)|\leq \varepsilon g(x) $$
Therefore
$$f(x)\approx_{x\rightarrow +\infty}g(x) $$
