Question

problems 7 - 8, use the properties of limits to evaluate. justify each step with a named rule. 7. $lim_{x
ightarrow - 3}(5x + 9)$ 8. $lim_{x
ightarrow 4}\frac{3x}{x - 2}$

Explanation:

Step1: Apply sum - rule of limits

The sum - rule of limits states that $\lim_{x
ightarrow a}(f(x)+g(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$. For $\lim_{x
ightarrow - 3}(5x + 9)$, we have $f(x)=5x$ and $g(x)=9$. So, $\lim_{x
ightarrow - 3}(5x + 9)=\lim_{x
ightarrow - 3}(5x)+\lim_{x
ightarrow - 3}(9)$.

Step2: Apply constant - multiple rule and constant rule

The constant - multiple rule of limits is $\lim_{x
ightarrow a}(cf(x))=c\lim_{x
ightarrow a}f(x)$ and the constant rule is $\lim_{x
ightarrow a}c = c$ (where $c$ is a constant). For $\lim_{x
ightarrow - 3}(5x)$, using the constant - multiple rule with $c = 5$ and $f(x)=x$, we get $5\lim_{x
ightarrow - 3}x$, and $\lim_{x
ightarrow - 3}9 = 9$. Also, $\lim_{x
ightarrow - 3}x=-3$ (by the direct - substitution property for the identity function $y = x$). Then $5\lim_{x
ightarrow - 3}x=5\times(-3)=-15$.

Step3: Calculate the final result

$\lim_{x
ightarrow - 3}(5x)+\lim_{x
ightarrow - 3}(9)=-15 + 9=-6$.

for problem 8:

Step1: Apply quotient - rule of limits

The quotient - rule of limits states that $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ (where $\lim_{x
ightarrow a}g(x)
eq0$). For $\lim_{x
ightarrow 4}\frac{3x}{x - 2}$, we have $f(x)=3x$ and $g(x)=x - 2$. So, $\lim_{x
ightarrow 4}\frac{3x}{x - 2}=\frac{\lim_{x
ightarrow 4}(3x)}{\lim_{x
ightarrow 4}(x - 2)}$.

Step2: Apply constant - multiple rule and difference - rule

For $\lim_{x
ightarrow 4}(3x)$, using the constant - multiple rule $\lim_{x
ightarrow a}(cf(x))=c\lim_{x
ightarrow a}f(x)$ with $c = 3$ and $f(x)=x$, we get $3\lim_{x
ightarrow 4}x$. And for $\lim_{x
ightarrow 4}(x - 2)$, using the difference - rule $\lim_{x
ightarrow a}(f(x)-g(x))=\lim_{x
ightarrow a}f(x)-\lim_{x
ightarrow a}g(x)$, we have $\lim_{x
ightarrow 4}x-\lim_{x
ightarrow 4}2$. Since $\lim_{x
ightarrow 4}x = 4$ (direct - substitution for $y = x$) and $\lim_{x
ightarrow 4}2 = 2$ (constant rule), $3\lim_{x
ightarrow 4}x=3\times4 = 12$ and $\lim_{x
ightarrow 4}x-\lim_{x
ightarrow 4}2=4 - 2=2$.

Step3: Calculate the final result

$\frac{\lim_{x
ightarrow 4}(3x)}{\lim_{x
ightarrow 4}(x - 2)}=\frac{12}{2}=6$.

Answer:

$-6$