## Calculus: Early Transcendentals 8th Edition

The statement is false. Here's an example to disprove it: $\lim\limits_{x\to0}f(x)=\lim\limits_{x\to0}\frac{x^2+5x+7}{x}=\frac{0^2+5\times0+7}{0}=\frac{7}{0}=\infty$ $\lim\limits_{x\to0}g(x)=\lim\limits_{x\to0}\frac{x^2+5x+6}{x}=\frac{0^2+5\times0+6}{0}=\frac{6}{0}=\infty$ However, we see that $\lim\limits_{x\to0}[f(x)-g(x)]=\lim\limits_{x\to0}\Big(\frac{x^2+5x+7}{x}-\frac{x^2+5x+6}{x}\Big)=\lim\limits_{x\to0}\Big(\frac{1}{x}\Big)=\frac{1}{0}=\infty\ne0$