Question

7–13. assume \\(\lim\limits_{x \to 1} f(x) = 8\\), \\(\lim\limits_{x \to 1} g(x) = 3\\), and \\(\lim\limits_{x \to 1} h(x) = 2\\). compute the following limits and state the limit laws used to justify your computations. 7. \\(\lim\limits_{x \to 1} (4f(x))\\) 8. \\(\lim\limits_{x \to 1} \frac{f(x)}{h(x)}\\) 9. \\(\lim\limits_{x \to 1} (f(x) - g(x))\\) 10. \\(\lim\limits_{x \to 1} (f(x)h(x))\\) 11. \\(\lim\limits_{x \to 1} \frac{f(x)}{g(x) - h(x)}\\) 12. \\(\lim\limits_{x \to 1} \sqrt3{f(x)g(x) + 3}\\)

Explanation:

Problem 7: $\lim\limits_{x\to 1}(4f(x))$

Step1: Use Constant Multiple Law

The Constant Multiple Law states that $\lim\limits_{x\to a}(cf(x)) = c\lim\limits_{x\to a}f(x)$ where $c$ is a constant. Here, $c = 4$, $a = 1$, and $\lim\limits_{x\to 1}f(x)=8$.
$\lim\limits_{x\to 1}(4f(x)) = 4\lim\limits_{x\to 1}f(x)$

Step2: Substitute the limit of $f(x)$

Substitute $\lim\limits_{x\to 1}f(x) = 8$ into the expression.
$4\times8 = 32$

Step1: Use Quotient Law

The Quotient Law states that $\lim\limits_{x\to a}\frac{f(x)}{h(x)}=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}h(x)}$ provided that $\lim\limits_{x\to a}h(x)
eq0$. Here, $a = 1$, $\lim\limits_{x\to 1}f(x)=8$, and $\lim\limits_{x\to 1}h(x)=2
eq0$.
$\lim\limits_{x\to 1}\frac{f(x)}{h(x)}=\frac{\lim\limits_{x\to 1}f(x)}{\lim\limits_{x\to 1}h(x)}$

Step2: Substitute the limits

Substitute $\lim\limits_{x\to 1}f(x) = 8$ and $\lim\limits_{x\to 1}h(x) = 2$ into the expression.
$\frac{8}{2}=4$

Step1: Use Difference Law

The Difference Law states that $\lim\limits_{x\to a}(f(x)-g(x))=\lim\limits_{x\to a}f(x)-\lim\limits_{x\to a}g(x)$. Here, $a = 1$, $\lim\limits_{x\to 1}f(x)=8$, and $\lim\limits_{x\to 1}g(x)=3$.
$\lim\limits_{x\to 1}(f(x)-g(x))=\lim\limits_{x\to 1}f(x)-\lim\limits_{x\to 1}g(x)$

Step2: Substitute the limits

Substitute $\lim\limits_{x\to 1}f(x) = 8$ and $\lim\limits_{x\to 1}g(x) = 3$ into the expression.
$8 - 3=5$

Answer:

$32$ (using Constant Multiple Law)

Problem 8: $\lim\limits_{x\to 1}\frac{f(x)}{h(x)}$