## Calculus: Early Transcendentals (2nd Edition)

$\lim_{x\to1}\dfrac{f(x)}{g(x)-h(x)}=8$
$\lim_{x\to1}\dfrac{f(x)}{g(x)-h(x)}$ It is known that $\lim_{x\to1}f(x)=8$ $,$ $\lim_{x\to1}g(x)=3$ and $\lim_{x\to1}h(x)=2$ Evaluate the limit by using the limit laws: $\lim_{x\to1}\dfrac{f(x)}{g(x)-h(x)}=...$ If the limit of the denominator is different from $0$, the limit of a quotient is the quotient of the limits of the numerator and the denominator: $...=\dfrac{\lim_{x\to1}f(x)}{\lim_{x\to1}[g(x)-h(x)]}=...$ The limit of a difference is the difference of the limits: $...=\dfrac{\lim_{x\to1}f(x)}{\lim_{x\to1}g(x)-\lim_{x\to1}h(x)}=...$ The limits indicated are known. Substitute them into the expression and simplify to evaluate the limit given: $...=\dfrac{8}{3-2}=\dfrac{8}{1}=8$