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