## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0}g(x)=4$$
$$\lim_{x\to0}\Big(\frac{4-g(x)}{x}\Big)=1$$ Here we cannot apply Quotient Rule because in the denominator, $\lim_{x\to0}x=0$. Instead, we would start with $g(x)$: $$g(x)=-(-g(x))=-(4-g(x)-4)=-\Big(\frac{4-g(x)}{x}\times x-4\Big)$$ Therefore, $$\lim_{x\to0}g(x)=\lim_{x\to0}\Big[-\Big(\frac{4-g(x)}{x}\times x-4\Big)\Big]$$ $$\lim_{x\to0}g(x)=-\Big[\lim_{x\to0}\Big(\frac{4-g(x)}{x}\Big)\times\lim_{x\to0}(x)-\lim_{x\to0}4\Big]$$ $$\lim_{x\to0}g(x)=-(1\times0-4)=-(-4)=4$$