## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x \to 1}\frac{x-3}{x^2+2x-4}=\frac{\lim\limits_{x \to 1}(x-3)}{\lim\limits_{x \to 1}(x^2+2x-4)}$$ Consider the denominator $$\lim\limits_{x \to 1}(x^2+2x-4)=1^2+2\times1-4=-1$$ Consdier the numerator $$\lim\limits_{x \to 1}(x-3)=1-3=-2$$ Therefore, both the limits exist. The limit of the denominator does not equal 0. So, it satisfies all the requirements of the quotient law, which states that $$\lim\limits_{x \to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}$$ In conclusion, the statement is true.