Answer
$\lim_{x\to1}\dfrac{f(x)}{g(x)-h(x)}=8$
Work Step by Step
$\lim_{x\to1}\dfrac{f(x)}{g(x)-h(x)}$
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:
$...=\dfrac{8}{3-2}=\dfrac{8}{1}=8$
