# What is the limit of $x/(x+\sin x)$ as $x$ approaches infinity?

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 $0$.

• Hint: $x-1\le x+\sin x\le x+1$. – David Mitra Jan 20 '15 at 18:32
• 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 $+\infty$ – 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 $-1\le\sin(x)\le1$.

• 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. – Ferris Jan 20 '15 at 18:55
• yes im since $-1\le \sin(x)\le 1$ is hold. – Dr. Sonnhard Graubner Jan 20 '15 at 18:56
• @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 $0$. – Erick Wong Jan 26 '17 at 5:41
• just use the squeeze theorem directly then... – The Great Duck Jun 10 '17 at 23:59
• 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. – 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.

• we call it the sandwich theorem in India! – Darshan Chaudhary Sep 13 '16 at 11:22
• My British maths teacher hates the name squeeze theorem and also calls it the sandwich theorem – 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.

• The tone of your post is pretty condescending. – 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 $1$.

Also

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 $|\sin(x)| < 1$.