The limit $\lim_{x\to \infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x}$ How to find the value of the limit:
$$ \lim_{x\to\infty}\frac{x-\frac{1}{2}\sin x}{x+\frac{1}{2}\sin x} $$
(l'Hopital not working here, right?)

Numerator and denominator are both $\sim x$, because $\lim_{x \to \infty} \frac{x \pm \frac 1 2 \sin x}{x}=1$, thus your limit is also 1.
Two functions $f$ and $g$ are equivalent at some point $a$ (possibly $\infty$), and you denote $f \sim g$, if: $f(x) = (1+\varepsilon(x))g(x)$ with $\varepsilon(x) \to 0$ for $x\to a$
Or if $g(x)$ is never 0, $\frac{f(x)}{g(x)} \to 1$.
Thus from the limit $\frac{\sin x}{x} \to 0$, you get immediately that $x-\frac 1 2 \sin x \sim x$, and similarly $x+\frac 1 2 \sin x \sim x$.
And since you can divide such "equivalents" provided the limit is not 0, you get the original limit from
$$\frac{x-\frac 1 2 \sin x}{x+\frac 1 2 \sin x} \sim \frac x x \sim 1$$

You can also write, for $x \to \infty$,
$$\frac{x-\frac 1 2\sin x}{x+\frac 1 2\sin x}=\frac{1-\frac 1 2\frac{\sin x}{x}}{1-\frac 1 2\frac{\sin x}{x}} \longrightarrow 1$$
The limit follows from the fact that $x \to \infty$ and $\sin x$ is bounded, thus $\frac{\sin x}{x} \to 0$.
