Let's solve the given limits step by step:
### Given Limits:
- [tex]\(\lim_{x \to a} f(x) = -2\)[/tex]
- [tex]\(\lim_{x \to a} g(x) = 0\)[/tex]
- [tex]\(\lim_{x \to a} h(x) = 27\)[/tex]
### Required Limits:
#### a. [tex]\(\lim_{x \to a} \sqrt[3]{h(x)}\)[/tex]
To find this limit, we simply take the cube root of the limit of [tex]\(h(x)\)[/tex]:
[tex]\[
\lim_{x \to a} \sqrt[3]{h(x)} = \sqrt[3]{\lim_{x \to a} h(x)} = \sqrt[3]{27} = 3
\][/tex]
#### b. [tex]\(\lim_{x \to a} \frac{g(x)}{f(x)}\)[/tex]
For this limit, we use the given limits:
[tex]\[
\lim_{x \to a} \frac{g(x)}{f(x)} = \frac{\lim_{x \to a} g(x)}{\lim_{x \to a} f(x)} = \frac{0}{-2} = 0
\][/tex]
#### c. [tex]\(\lim_{x \to a} \frac{f(x)}{g(x)}\)[/tex]
Since [tex]\( \lim_{x \to a} g(x) = 0 \)[/tex], the denominator is zero, meaning this limit does not exist (or is undefined), because division by zero is undefined.
#### d. [tex]\(\lim_{x \to a} \frac{3 f(x)}{h(x) + 9 f(x)}\)[/tex]
First, we substitute the given limits into the expression:
[tex]\[
\text{Numerator: } 3 \lim_{x \to a} f(x) = 3(-2) = -6
\][/tex]
[tex]\[
\text{Denominator: } \lim_{x \to a} (h(x) + 9 f(x)) = \lim_{x \to a} h(x) + 9 \lim_{x \to a} f(x) = 27 + 9(-2) = 27 - 18 = 9
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \to a} \frac{3 f(x)}{h(x) + 9 f(x)} = \frac{-6}{9} = -\frac{2}{3}
\][/tex]
### Additional Limits:
#### a. [tex]\(\lim_{x \to 0} (7x^2 - 5x + 1)\)[/tex]
To evaluate this limit, we substitute [tex]\( x = 0 \)[/tex] into the polynomial expression:
[tex]\[
7(0)^2 - 5(0) + 1 = 0 - 0 + 1 = 1
\][/tex]
#### b. [tex]\(\lim_{x \to 2} \frac{2x - 3}{x + 5}\)[/tex]
To find this limit, we substitute [tex]\( x = 2 \)[/tex] into the rational expression:
[tex]\[
\frac{2(2) - 3}{2 + 5} = \frac{4 - 3}{7} = \frac{1}{7}
\][/tex]
### Summary of the results:
a. [tex]\( \lim_{x \to a} \sqrt[3]{h(x)} = 3 \)[/tex]
b. [tex]\( \lim_{x \to a} \frac{g(x)}{f(x)} = 0 \)[/tex]
c. [tex]\( \lim_{x \to a} \frac{f(x)}{g(x)} \)[/tex] is undefined.
d. [tex]\( \lim_{x \to a} \frac{3 f(x)}{h(x) + 9 f(x)} = -\frac{2}{3} \)[/tex]
### Evaluate Additional Limits:
a. [tex]\( \lim_{x \to 0} (7x^2 - 5x + 1) = 1 \)[/tex]
b. [tex]\( \lim_{x \to 2} \frac{2x - 3}{x + 5} = \frac{1}{7} \)[/tex]
These results complete the solutions to the provided exercise.
