Need a hint to evaluate $\lim_{x\to +\infty}{x-\sin x \over x+\sin x}$

Question

I Need a hint to evaluate $\lim\limits_{x\to +\infty} \dfrac{x-\sin x}{x+\sin x}$ I was doing the following $$-1\le \sin x \le1$$ $$1 \ge -\sin x \ge -1$$ $$ {x+1 \over x+\sin x} \ge {x-\sin x \over x + \sin x} \ge {x-1 \over x+\sin x}$$ Now I think, by Squeeze Theorem, that $\lim_{x\to +\infty}{x-\sin x \over x+\sin x}=1$

Is there other way to solve it without L'Hopital's rule?

Answer 1

$x > 2,$ $$ 1 - \frac{2}{x+1} = \frac{x-1}{x+1} \leq \frac{x-\sin x}{x+\sin x} \leq \frac{x+1}{x-1} = 1 + \frac{2}{x-1} $$

Answer 2

HINT:

$$\frac{x-\sin x}{x+\sin x}=\frac{1-\frac{\sin x}x}{1+\frac{\sin x}x}$$

Answer 3

A bit overkill, but gives the second-order term as well:

If you are familiar with Taylor expansions? Rewriting your expression as $$ \frac{x-\sin x}{x+\sin x} = \frac{1-\frac{\sin x}{x}}{1+\frac{\sin x}{x}} $$ yand observing that $\frac{\sin x}{x}\xrightarrow[x\to\infty]{}0$, you can use the fact that $\frac{1}{1+y} = 1-y + o(y)$ when $y\to 0$.

You will get $$ \frac{x-\sin x}{x+\sin x} = (1-\frac{\sin x}{x})(1-\frac{\sin x}{x}+o\left(\frac{1}{x}\right)) = 1-2\frac{\sin x}{x}+o\left(\frac{1}{x}\right) $$ which gives the limit (and a bit more).

Answer 4

It's shorter with equivalents, as often: $x\pm \sin x\sim_\infty x$, hence: $$\frac{x-\sin x}{x+\sin x}\sim_\infty\frac xx=1. $$