# Limit with sin indeterminate

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 $3$.

• Divide the numerator and denominator by $x$ – Minz May 17 '17 at 6:08
• How did you get $-1$? If you show us, may be we can help find the mistake. – Paramanand Singh May 17 '17 at 10:19
• I wrote : $$\lim_{x\to\infty} \frac{3(x+\sin{x})-4\sin{x}}{x+\sin{x}}$$ and then I get something like: $$\lim_{x\to\infty} 3- \frac{4\sin{x}}{x+\sin{x}}$$ I used l'hopital twice for the second one and I would get 7 not 1(I forgot a sign when I derived).What is wrong when I split it like this? – Lola May 17 '17 at 13:58
• L'hopital's rule only works if you have an indeterminate limit. The second term you got above is not interdeterminate. Plug in infinity and see what you get. – David Stanley May 17 '17 at 14:39
• You don't need any fancy tricks for this one. Just imagine $x$ getting really huge. What's going to happen to the numerator? The denominator? Will the extra $\sin x$ terms make much difference? – Kevin May 17 '17 at 16:51

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 $3$.

• >For the second expression, the numerator is bounded while the denominator is not, so It's not enough that denominator is not bounded (if it was smth like 1/(xsinx + 1) limit would not exist. – RiaD May 17 '17 at 21:00
• @RiaD Fair enough, I fixed it to be more precise. The key is that, in the denominator, the $x$ term goes to infinity while the sine term added to it remains bounded. – Deepak May 18 '17 at 0:45

Hint: divide numerator and denominator by $x$.

One has

$$0\leq\left\vert{\sin{x}\over x}\right\vert\leq{1\over |x|}$$

And so ${\sin{x}\over x}\to 0$ as $x\to\infty$. Now in the fraction divide the numerator and denominator by $x$ to get

$${3-{\sin{x}\over x}\over 1-{\sin{x}\over x}}\to 3$$

• Is there a way to use l'Hopital here?Because I tried to split them but it didn't work – Lola May 17 '17 at 6:29
• $\frac{3-\cos x}{1+\cos x}$ has poles at $(2k+1)\pi$, and it takes the value $1$ at $2k\pi$. L'Hospital doesn't work for $\frac{3x-\sin x}{x+\sin x}$ directly. – Daniel Fischer May 17 '17 at 9:59
• A couple of things. First, please fix the sign error in your limit - the denominator must have a plus sign. Second, L'Hopital is not applicable here because the limit of the ratio (after taking derivatives) does not exist. Even though $\cos x$ is bounded, the rational expression has no defined limit as $x \to \infty$. Remember that LHR is only applicable if a set of criteria are met and one of those criteria is that the limit of the "reduced" expression also exist. In this case, it does not, so LHR is not useful. – Deepak May 17 '17 at 10:38

$$\lim_{x\to \infty} \frac{3x-\sin x}{x+\sin x}=\lim_{x\to \infty} \frac{3-\frac{\sin x}{x}}{1+\frac{\sin x}{x}}$$

and $\sin x$ is bounded in $\mathbb{R}$

• Bounded in $\mathbb{R}$, to be precise ;-) – ComplexFlo May 17 '17 at 6:09
• yeah, now i have edited. – Arun May 17 '17 at 6:16