$\lim_{x\rightarrow \infty} \dfrac{x-\sin x}{x+\cos x}$
$= \lim_{x\rightarrow \infty} \dfrac{x(1-\frac{\sin x}{x})}{x(1+\frac{\cos x}{x})}$
$= \lim_{x\rightarrow \infty} \dfrac{1-\frac{\sin x}{x}}{1+\frac{\cos x}{x}}$
now to calculate values of $\frac{\sin x}{x}$ and $\frac{\cos x}{x}$ we use Squeezing Theorem.
$-1\leq \sin x\leq {+1} \\ \dfrac{-1}{x}\leq \dfrac{\sin x}{x}\leq \dfrac{+1}{x}$
$-1\leq \cos x\leq {+1} \\ \dfrac{-1}{x}\leq \dfrac{\cos x}{x}\leq \dfrac{+1}{x}$
now as $x \rightarrow \infty$ we get $\frac{1}{x} \rightarrow 0$, this implies that:
$0\leq \dfrac{\sin x}{x} \leq 0 \\ 0\leq \dfrac{\cos x}{x} \leq 0$
Hence,
$\lim_{x\rightarrow \infty} \frac{x-\sin x}{x+\cos x} $
$= \lim_{x\rightarrow \infty} \frac{1-\frac{\sin x}{x}}{1+\frac{\cos x}{x}}$
$= \lim_{x\rightarrow \infty} \frac{1-0}{1+0}\\ = 1$
answer = option A