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 $0$.
 A: 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

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

A: 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.
A: 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 $-1\le\sin(x)\le1$.
A: $-1\leq\sin x\leq 1$ 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.
A: Since $x\to+\infty$ and $-1\le\sin(x)\le1$ for all $x\in\Bbb R$, we have that $\sin(x)=o(x)$ and
$$
\lim_{x\to\infty}\frac{x}{x+\sin x}=
\lim_{x\to\infty}\frac{x}{x+o(x)}=
\lim_{x\to\infty}\frac x x=1.
$$
A: $\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 $.
