Answer
$$\lim_{x\to\infty}h(x)=\lim_{x\to-\infty}h(x)=\frac{9}{2}$$
Work Step by Step
To solve these exercises, 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) $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{9x^4+x}{2x^4+5x^2-x+6}$$
The highest power of $x$ in the denominator here is $x^4$, so we divide both numerator and denominator by $x^4$: $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{9+\frac{1}{x^3}}{2+\frac{5}{x^2}-\frac{1}{x^3}+\frac{6}{x^4}}$$ $$\lim_{x\to\infty}h(x)=\frac{9+0}{2+0-0+0}=\frac{9}{2}$$
(b) $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{9x^4+x}{2x^4+5x^2-x+6}$$
Again, we divide both numerator and denominator by $x^4$: $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{9+\frac{1}{x^3}}{2+\frac{5}{x^2}-\frac{1}{x^3}+\frac{6}{x^4}}$$ $$\lim_{x\to-\infty}h(x)=\frac{9+0}{2+0-0+0}=\frac{9}{2}$$
