Answer
$$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\frac{2}{5}$$
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) $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{2x+3}{5x+7}$$
The highest power of $x$ in the denominator here is $x$, so we divide both numerator and denominator by $x$: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{2+\frac{3}{x}}{5+\frac{7}{x}}=\frac{\lim_{x\to\infty}2+\lim_{x\to\infty}\frac{3}{x}}{\lim_{x\to\infty}5+\lim_{x\to\infty}\frac{7}{x}}$$ $$\lim_{x\to\infty}f(x)=\frac{2+0}{5+0}=\frac{2}{5}$$
(b) $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{2x+3}{5x+7}$$
Again, we divide both numerator and denominator by $x$: $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{2+\frac{3}{x}}{5+\frac{7}{x}}=\frac{\lim_{x\to-\infty}2+\lim_{x\to-\infty}\frac{3}{x}}{\lim_{x\to-\infty}5+\lim_{x\to-\infty}\frac{7}{x}}$$ $$\lim_{x\to-\infty}f(x)=\frac{2+0}{5+0}=\frac{2}{5}$$
