TITLE: Limit with sin indeterminate
QUESTION [6 upvotes]: How do I calculate the following limit
$$\lim_{x\to\infty} \frac{3x-\sin x}{x+\sin x}$$
It's an indeterminate limit but how can I solve it? Does it help if I split it?The answer I got is $-1$ but it's $3$.

REPLY [9 votes]: Just for the sake of variety, without dividing by $x$,
$$\frac{3x - \sin x}{x + \sin x} = \frac{3(x+ \sin x) - 4\sin x}{x + \sin x} = 3 - \frac{4\sin x}{x + \sin x}$$
For the second expression, the numerator is bounded while the denominator tends to positive infinity, so $\displaystyle \lim_{x \to \infty}\frac{4\sin x}{x + \sin x} = 0$, hence the original limit is $3$.