I am trying to determine

$$\lim_{x \to \infty} \frac{x}{x+ \sin x} $$

I can't use here the remarkable limit (I don't know if I translated that correctly) $ \lim_{x\to 0} \frac{\sin x}{x}=1$ because $x$ approaches infinity, not $0$.

  • 5
    $\begingroup$ Hint: $x-1\le x+\sin x\le x+1$. $\endgroup$ – David Mitra Jan 20 '15 at 18:32
  • $\begingroup$ Keep in mind that $x+\sin{x}\gt x-1$ this means that $\frac{x}{x+\sin{x}}\lt \frac{x}{x-1}$. This should help for the limit at $+\infty$ $\endgroup$ – marwalix Jan 20 '15 at 18:36

Assume $x\neq 0$ and divide the given term by $x$ to get the form $\frac{1}{1+\frac{\sin(x)}{x}}$. This clearly tends to $1$ as $x\rightarrow\infty$ since $-1\le\sin(x)\le1$.

  • $\begingroup$ Are you sure lim[(sinx)/x] = 0 when x aproaches infinity ? I mean it's obvious it's 0 because you divide a number between -1 and +1 with something that approaches infinity, but we now study limits and we were not told that. $\endgroup$ – Ferris Jan 20 '15 at 18:55
  • 2
    $\begingroup$ yes im since $-1\le \sin(x)\le 1$ is hold. $\endgroup$ – Dr. Sonnhard Graubner Jan 20 '15 at 18:56
  • 8
    $\begingroup$ @Ferris The most common way to prove this at the beginner level is called the squeeze theorem: $-\frac1x \le \frac{\sin x}{x} \le \frac1x$ and both the left and right sides converge to $0$, which forces the middle term to also converge to $0$. $\endgroup$ – Erick Wong Jan 26 '17 at 5:41
  • $\begingroup$ just use the squeeze theorem directly then... $\endgroup$ – The Great Duck Jun 10 '17 at 23:59
  • $\begingroup$ agreed, one can simply make the case that $$ \frac{1}{1+ \frac{-1}{x}} \leq f(x) \leq \frac{1}{1+ \frac{1}{x}}$$ and apply squeeze theorem. $\endgroup$ – Wesley Strik Jan 27 at 11:30

$-1\leq\sin x\leq 1$ so $$\frac{x}{x+1}\leq\frac{x}{x+\sin x}\leq \frac{x}{x-1}$$
Can you show that the limit of $x/(x+1)$ and of $x/(x-1)$ are both $1$? Then use the Squeeze Theorem.

  • 3
    $\begingroup$ we call it the sandwich theorem in India! $\endgroup$ – Darshan Chaudhary Sep 13 '16 at 11:22
  • $\begingroup$ My British maths teacher hates the name squeeze theorem and also calls it the sandwich theorem $\endgroup$ – theonlygusti Sep 24 '17 at 15:05

Since $x$ is positive and non-zero as $x\to\infty$, we have

$$ -1\leq\sin x\leq 1$$ $$ -\frac{1}{x}\leq \frac{\sin x}{x} \leq \frac{1}{x}$$ $$ -\lim\limits_{x\to\infty}\frac{1}{x}\leq \lim\limits_{x\to\infty}\frac{\sin x}{x} \leq\lim\limits_{x\to\infty}\frac{1}{x}$$ $$ 0\leq \lim\limits_{x\to\infty}\frac{\sin x}{x} \leq 0$$

Therefore by the squeeze theorem,

$$\lim\limits_{x \to \infty} \frac{\sin x}{x}=0$$

So now we have

$$\lim\limits_{x \to \infty} \frac{x}{x+ \sin x} = \lim\limits_{x \to \infty} \frac{1}{1+ \frac{\sin x}{x}} $$ $$= \frac{1}{1+ \lim\limits_{x \to \infty} \frac{\sin x}{x}} =\frac{1}{1+0}=1$$


Think of $\displaystyle\frac{\text{1 trillion}}{(\text{1 trillion}) + \sin(\text{1 trillion})} = \frac{\text{1 trillion}}{(\text{1 trillion}) + (\text{a number between $1$ and $-1$})}$.

If you understand that, then you will see how that leads to the answer.

To do it in a more precise way, squeeze: $$ \frac x {x+1} \le \frac x {x+\sin x} \le \frac x {x-1} $$ Squeezing is usually something to be considered when dealing with sines.

  • 19
    $\begingroup$ The tone of your post is pretty condescending. $\endgroup$ – Cameron Williams Sep 2 '15 at 18:34

Since $x\to+\infty$ and $-1\le\sin(x)\le1$ for all $x\in\Bbb R$, we have that $\sin(x)=o(x)$ and

$$ \lim_{x\to\infty}\frac{x}{x+\sin x}= \lim_{x\to\infty}\frac{x}{x+o(x)}= \lim_{x\to\infty}\frac x x=1. $$


$\sin x $ is bounded within $\pm 1$, $x$ are unbounded and equal, so it tends to $1$.


it is known $ \frac{\sin x }{x} \rightarrow 0 $ individually.

So $\dfrac{1}{1+\dfrac{\sin(x)}{x}}$.

must tend to $1$ as $x\rightarrow\infty$ since $|\sin(x)| < 1 $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.