#### Answer

$$\lim_{x\to\infty}\frac{2+\sqrt x}{2-\sqrt x}=-1$$

#### Work Step by Step

We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator.
Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to\infty}\frac{2+\sqrt x}{2-\sqrt x}$$
The highest power of $x$ in the denominator here is $\sqrt x$, so we divide both numerator and denominator by $\sqrt x$: $$A=\lim_{x\to\infty}\frac{\frac{2}{\sqrt x}+1}{\frac{2}{\sqrt x}-1}=\lim_{x\to\infty}\frac{\frac{2}{x^{1/2}}+1}{\frac{2}{x^{1/2}}-1}=\frac{\lim_{x\to\infty}\frac{2}{x^{1/2}}+\lim_{x\to\infty}1}{\lim_{x\to\infty}\frac{2}{x^{1/2}}-\lim_{x\to\infty}1}$$
$$A=\frac{0+1}{0-1}=\frac{1}{-1}=-1$$