Answer
$$\lim_{x\to\infty}\frac{x+\sin x+2\sqrt x}{x+\sin x}=1$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to\infty}\frac{x+\sin x+2\sqrt x}{x+\sin x}$$
Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$ here:
$$A=\lim_{x\to\infty}\frac{1+\frac{\sin x}{x}+\frac{2}{\sqrt x}}{1+\frac{\sin x}{x}}$$
$$A=\frac{1+\lim_{x\to\infty}\frac{\sin x}{x}+0}{1+\lim_{x\to\infty}\frac{\sin x}{x}}=\frac{1+\lim_{x\to\infty}\frac{\sin x}{x}}{1+\lim_{x\to\infty}\frac{\sin x}{x}}$$
We know that $-1\le\sin x\le1$, hence $\frac{-1}{x}\le\frac{\sin x}{x}\le\frac{1}{x}$ (because $x\to\infty$, we assume $x\gt0$ here)
Furthermore, $\lim_{x\to\infty}-1/x=\lim_{x\to\infty}1/x=0$
Therefore, according to Sandwich Theorem, $\lim_{x\to\infty}(\sin x/x)=0$
$$A=\frac{1+0}{1+0}=1$$
