Answer
$$\lim_{x\to\infty}h(x)=\lim_{x\to-\infty}h(x)=7$$
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) $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{7x^3}{x^3-3x^2+6x}$$
The highest power of $x$ in the denominator here is $x^3$, so we divide both numerator and denominator by $x^3$: $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{7}{1-\frac{3}{x}+\frac{6}{x^2}}$$ $$\lim_{x\to\infty}h(x)=\frac{7}{1-0+0}=\frac{7}{1}=7$$
(b) $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{7x^3}{x^3-3x^2+6x}$$
Again, we divide both numerator and denominator by $x^3$: $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{7}{1-\frac{3}{x}+\frac{6}{x^2}}$$ $$\lim_{x\to-\infty}h(x)=\frac{7}{1-0+0}=\frac{7}{1}=7$$
