Answer
a. $y=-1$
b. $y=1$
c. $y=\pm1$
d. $y=\frac{1}{3}$
Work Step by Step
To determine the horizontal asymptotes, we evaluate the limits of the function when $x\to\pm\infty$
a. $\lim_{x\to-\infty}\frac{1-x^2}{x^2+1}=\lim_{x\to-\infty}\frac{1/x^2-1}{1+1/x^2}=-1$, $\lim_{x\to\infty}\frac{1-x^2}{x^2+1}=\lim_{x\to\infty}\frac{1/x^2-1}{1+1/x^2}=-1$, thus $y=-1$ is the horizontal asymptote.
b. $x\geq0$, $\lim_{x\to\infty}\frac{\sqrt x+4}{\sqrt {x+4}}=\lim_{x\to\infty}\frac{1+4/\sqrt x}{\sqrt {1+4/x}}=1$, thus $y=1$ is the horizontal asymptote.
c. $\lim_{x\to-\infty}\frac{\sqrt {x^2+4}}{x}=\lim_{x\to-\infty}-\sqrt {1+4/x^2}=-1$, $\lim_{x\to\infty}\frac{\sqrt {x^2+4}}{x}=\lim_{x\to\infty}\sqrt {1+4/x^2}=1$, thus $y=\pm1$ are the horizontal asymptotes.
d. $\lim_{x\to\pm\infty}\sqrt {\frac{x^2+9}{9x^2+1}}=\lim_{x\to\pm\infty}\sqrt {\frac{1+9/x^2}{9+1/x^2}}=\frac{1}{3}$, thus $y=\frac{1}{3}$ is the horizontal asymptote.
