Answer
(a) $$\lim_{x\to\infty}h(x)=-\infty$$
(b) $$\lim_{x\to-\infty}h(x)=\infty$$
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) $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{5x^8-2x^3+9}{3+x-4x^5}$$
The highest power of $x$ in the denominator here is $x^5$, so we divide both numerator and denominator by $x^5$: $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{5x^3-\frac{2}{x^2}+\frac{9}{x^5}}{\frac{3}{x^5}+\frac{1}{x^4}-4}$$ $$\lim_{x\to\infty}h(x)=\frac{\lim_{x\to\infty}(5x^3)-0+0}{0+0-4}=\frac{\lim_{x\to\infty}(5x^3)}{-4}=-\lim_{x\to\infty}(\frac{5x^3}{4})$$
As $x\to\infty$, $(x^3)$ approaches $\infty$, so $(5/4)x^3$ will approach $\infty$ as well. Therefore, $$\lim_{x\to\infty}h(x)=-\lim_{x\to\infty}\Big(\frac{5x^3}{4}\Big)=-\infty$$
(b) $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{5x^8-2x^3+9}{3+x-4x^5}$$
Again, we divide both numerator and denominator by $x^5$: $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{5x^3-\frac{2}{x^2}+\frac{9}{x^5}}{\frac{3}{x^5}+\frac{1}{x^4}-4}$$ $$\lim_{x\to-\infty}h(x)=\frac{\lim_{x\to-\infty}(5x^3)-0+0}{0+0-4}=\frac{\lim_{x\to-\infty}(5x^3)}{-4}=-\lim_{x\to-\infty}(\frac{5x^3}{4})$$
As $x\to-\infty$, $(x^3)$ approaches $-\infty$, so $(5/4)x^3$ will approach $-\infty$ as well. Therefore, $$\lim_{x\to-\infty}h(x)=-\lim_{x\to-\infty}\Big(\frac{5x^3}{4}\Big)=-(-\infty)=\infty$$
