Answer
The equations of the horizontal asymptotes of the functions are:
a) $y=-1$
b) $y=1$
c) $y=-1$ and $y=1$
d) $y=1/3$
Work Step by Step
a) $$y=\frac{1-x^2}{x^2+1}$$ $$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\frac{1-x^2}{x^2+1}$$
Divide both numerator and denominator by $x^2$:
$$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\frac{\frac{1}{x^2}-1}{1+\frac{1}{x^2}}=\frac{0-1}{1+0}=\frac{-1}{1}=-1$$
According to definition, $y=-1$ is the horizontal asymptote of the function.
b) $$y=\frac{\sqrt x+4}{\sqrt{x+4}}$$
- Domain: $[0,\infty)$
So we only examine limits of $y$ as $x\to\infty$:
$$\lim_{x\to\infty}y=\lim_{x\to\infty}\frac{\sqrt x+4}{\sqrt{x+4}}$$
Divide both numerator and denominator by $\sqrt x$:
$$\lim_{x\to\infty}y=\lim_{x\to\infty}\frac{1+\frac{4}{\sqrt x}}{\frac{\sqrt{x+4}}{\sqrt x}}=\lim_{x\to\infty}\frac{1+\frac{4}{\sqrt x}}{\sqrt{1+\frac{4}{x}}}$$ $$\lim_{x\to\infty}y=\frac{1+0}{\sqrt{1+0}}=\frac{1}{\sqrt1}=1$$
According to definition, $y=1$ is the horizontal asymptote of the function.
c) $$g(x)=\frac{\sqrt{x^2+4}}{x}$$ $$\lim_{x\to\pm\infty}g(x)=\lim_{x\to\pm\infty}\frac{\sqrt{x^2+4}}{x}$$
- For $x\to\infty$: we assume $x\gt0$, and $x=|x|=\sqrt{x^2}$ $$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}\frac{\sqrt{x^2+4}}{\sqrt{x^2}}=\lim_{x\to\infty}\sqrt{1+\frac{4}{x^2}}=\sqrt{1+0}=1$$
- For $x\to-\infty$: we assume $x\lt0$, and $x=-|x|=-\sqrt{x^2}$ $$\lim_{x\to-\infty}g(x)=\lim_{x\to-\infty}\frac{\sqrt{x^2+4}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1+\frac{4}{x^2}}=-\sqrt{1+0}=-1$$
According to definition, $y=1$ and $y=-1$ are the horizontal asymptotes of the function.
d) $$y=\sqrt{\frac{x^2+9}{9x^2+1}}$$ $$\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\sqrt{\frac{x^2+9}{9x^2+1}}=\sqrt{\lim_{x\to\pm\infty}\frac{x^2+9}{9x^2+1}}$$
Divide both numerator and denominator by $x^2$:
$$\lim_{x\to\pm\infty}y=\sqrt{\lim_{x\to\pm\infty}\frac{1+\frac{9}{x^2}}{9+\frac{1}{x^2}}}=\sqrt{\frac{1+0}{9+0}}=\sqrt{\frac{1}{9}}=\frac{1}{3}$$
According to definition, $y=1/3$ is the horizontal asymptote of the function.