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