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

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

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

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

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