Calculating $\lim \limits_{x \to \infty} \frac{x+\frac12\cos x}{x-\frac12\sin x}$ using the sandwich theorem Calculating $\lim \limits_{x \to \infty} \dfrac{x+\frac12\cos x}{x-\frac12\sin x}$
Correct me if I'm wrong:
$\cos x$ and $\sin x$ are bounded so that 
$$|\cos x|\le 1,\qquad |\sin x|\le1$$
Therefore I can say:
$$
\frac{x-\frac12}{x+\frac12}\le
\frac{x+\frac12\cos x}{x-\frac12\sin x}\le 
\frac{x+\frac12}{x-\frac12}
$$
the limits of the left and right side are equal to 1, therefore the the limit I'm looking for is also equal to 1 . The answer is correct, but what I'm not sure is
$$
\frac{x-\frac12}{x+\frac12}\le\frac{x+\frac12\cos x}{x-\frac12\sin x}
$$
was this step correct?

Just to make more explicit the steps, we can assume $x>1$, so $x-\frac{1}{2}\sin x>0$ and $x+\frac{1}{2}\cos x>0$.
From $|\sin x|<1$, we get
$$
x-\frac{1}{2}\le x-\frac{1}{2}\sin x\le x+\frac{1}{2}
$$
Similarly,
$$
x-\frac{1}{2}\le x+\frac{1}{2}\cos x\le x+\frac{1}{2}
$$
Since all terms are positive, from
$$
x-\frac{1}{2}\le x-\frac{1}{2}\sin x
\quad\text{and}\quad
x-\frac{1}{2}\le x+\frac{1}{2}\cos x
$$
we obtain
$$
\frac{x-\frac{1}{2}}{x+\frac{1}{2}}\le
\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}
$$
Similarly,
$$
\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}\le
\frac{x+\frac{1}{2}}{x-\frac{1}{2}}
$$
and the sandwich theorem allows to conclude.
There is a different and perhaps simpler way to prove the result. It is not restrictive to assume $x>0$.
Since $|\cos x|\le1$, we have
$$
-\frac{1}{2x}\le\frac{\cos x}{2x}\le\frac{1}{2x}
$$
implying $\lim_{x\to\infty}\frac{\cos x}{2x}=0$ by the sandwich theorem. Similarly, $\lim_{x\to\infty}\frac{\sin x}{2x}=0$, so
$$
\lim_{x\to\infty}\frac{x+\frac{1}{2}\cos x}{x-\frac{1}{2}\sin x}
=
\lim_{x\to\infty}\frac{1+\dfrac{\cos x}{2x}}{1-\dfrac{\sin x}{2x}}
=\frac{1+0}{1+0}=1
$$
