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 $0$.

  • 6
    $\begingroup$ Hint: $x-1\le x+\sin x\le x+1$. $\endgroup$ Jan 20, 2015 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, 2015 at 18:36

6 Answers 6


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 $-1\le\sin(x)\le1$.

  • 1
    $\begingroup$ 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, 2015 at 18:55
  • 3
    $\begingroup$ yes im since $-1\le \sin(x)\le 1$ is hold. $\endgroup$ Jan 20, 2015 at 18:56
  • 13
    $\begingroup$ @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, 2017 at 5:41
  • $\begingroup$ 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. $\endgroup$
    – user459879
    Jan 27, 2019 at 11:30
  • $\begingroup$ @WesleyStrik Oughtn't your $\le$ be $\ge$? $\endgroup$
    – user53259
    Apr 23, 2019 at 3:28

$-1\leq\sin x\leq 1$ 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.

  • 7
    $\begingroup$ we call it the sandwich theorem in India! $\endgroup$ Sep 13, 2016 at 11:22
  • 1
    $\begingroup$ My British maths teacher hates the name squeeze theorem and also calls it the sandwich theorem $\endgroup$ Sep 24, 2017 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.

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    $\begingroup$ The tone of your post is pretty condescending. $\endgroup$ Sep 2, 2015 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 $.


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