How to solve limits with Taylor expansion?

Question

I'm in trouble with Taylor series..... how can I solve limits without Bernoulli-de L'Hôpital method?? For example, $$\lim_{x \to +\infty} \frac{x-\sin{x}}{2x+\sin{x}}.$$ The answer, if I'm not wrong is $\frac{1}{2}$... How, can I show that? Thanks.

Answer 1

$$\lim_{x \to +\infty} \frac{x-\sin{x}}{2x+\sin{x}}=\lim_{x \to +\infty} \frac{\frac xx-\frac {\sin{x}}{x}}{\frac {2x}{x}+\frac{\sin{x}}{x}}=\lim_{x \to +\infty} \frac{1-\frac {\sin{x}}{x}}{2+\frac{\sin{x}}{x}}$$

Now use $$-\frac{1}{x}\le\frac{\sin x}{x}\le\frac{1}{x}$$

Use sandwich theorem to show $$\lim_{x\to \infty}\frac{\sin x}{x}=0$$

$\text{Game Over!}$

Answer 2

Hint

Just rewrite $$\frac{x-\sin{x}}{2x+\sin{x}}=\frac{1-\frac{\sin{x}}{x}}{2+\frac{\sin{x}}{x}}$$ and take into account that $\sin(x)$ is bounded between $-1$ and $1$ and that $x$ is going to $\infty$.

I am sure that you can take from here.