Answer
(a) Quotient Rule.
(b) Numerator: Root Rule - Denominator: Product Rule
(c) Numerator: Constant Multiple Rule - Denominator: Difference Rule
Work Step by Step
(a) $$\lim_{x\to1}\frac{\sqrt{5h(x)}}{p(x)(4-r(x))}=\frac{\lim_{x\to1}\sqrt{5h(x)}}{\lim_{x\to1}(p(x)(4-r(x)))}$$
Here, $\lim_{x\to c}\frac{A}{B}=\frac{\lim_{x\to c}A}{\lim_{x\to c}B}$. This is Quotient Rule.
(b) $$\frac{\lim_{x\to1}\sqrt{5h(x)}}{\lim_{x\to1}(p(x)(4-r(x)))}=\frac{\sqrt{\lim_{x\to1}5h(x)}}{\Big(\lim_{x\to1}p(x)\Big)\Big(\lim_{x\to1}(4-r(x))\Big)}$$
- Numerator: $\lim_{x\to c}\sqrt[n]A=\sqrt[n]{\lim_{x\to c}A}$. This is Root Rule.
- Denominator: $\lim_{x\to c}(AB)=\lim_{x\to c}A\times\lim_{x\to c}B$. This is Product Rule.
(c) $$\frac{\sqrt{\lim_{x\to1}5h(x)}}{\Big(\lim_{x\to1}p(x)\Big)\Big(\lim_{x\to1}(4-r(x))\Big)}=\frac{\sqrt{5\lim_{x\to1}h(x)}}{\Big(\lim_{x\to1}p(x)\Big)\Big(\lim_{x\to1}4-\lim_{x\to1}r(x)\Big)}$$
- Numerator: $\lim_{x\to c}(kA)=k\lim_{x\to c}A$. This is Constant Multiple Rule.
- Denominator: Look at $\lim_{x\to1}(4-r(x))=\lim_{x\to1}4-\lim_{x\to1}r(x)$. This is Difference Rule.
