## Calculus: Early Transcendentals 8th Edition

This statement is false. Here's an example to disprove: $\lim\limits_{x\to5}f(x)=\lim\limits_{x\to5}(x^2-5x)=5^2-5\times5=0$ $\lim\limits_{x\to5}g(x)=\lim\limits_{x\to5}(x-5)=5-5=0$ However, $$\lim\limits_{x\to5}\frac{x^2-5x}{x-5}$$$$=\lim\limits_{x\to5}\frac{x(x-5)}{x-5}$$$$=\lim\limits_{x\to5}x$$$$=5$$ That means $\lim\limits_{x\to5}\frac{x^2-5x}{x-5}$ does exist.