5. Consider $$ f(x)=\frac{4-x}{3+x} $$. a. What are the asymptotes? b. What is the $$ \lim _{x \rightarrow-3^{+}} f(x) $$ ? c. What is the $$ \lim _{x \rightarrow-3^{-}} f(x) $$ ? d. What is the $$ \lim _{x \rightarrow \infty} f(x) $$ ?

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Let's analyze the function $$ f(x) = \frac{4 - x}{3 + x} $$ step by step for each part of the question.

### a. Asymptotes

**Vertical Asymptote:**
Vertical asymptotes occur where the denominator is zero (and the numerator is not zero).

Set the denominator equal to zero:
$$
3 + x = 0 \quad \Rightarrow \quad x = -3
$$
**Vertical Asymptote:** $$ x = -3 $$

**Horizontal Asymptote:**
For rational functions where the degree of the numerator and the denominator are the same, the horizontal asymptote is the ratio of the leading coefficients.

Here, both numerator and denominator are degree 1:
$$
\text{Leading coefficient of numerator} = -1 \quad (\text{from } -x)
$$
$$
\text{Leading coefficient of denominator} = 1 \quad (\text{from } x)
$$
$$
\text{Horizontal Asymptote}: y = \frac{-1}{1} = -1
$$

**Horizontal Asymptote:** $$ y = -1 $$

### b. $$ \lim_{{x \to -3^{+}}} f(x) $$

We evaluate the limit as $$ x $$ approaches $$-3$$ from the right (values greater than $$-3$$).

$$
f(x) = \frac{4 - x}{3 + x}
$$

As $$ x \to -3^+ $$:
- **Numerator:** $$ 4 - (-3) = 7 $$ (approaches 7)
- **Denominator:** $$ 3 + (-3^+) $$ approaches $$ 0^+ $$ (a small positive number)

$$
\lim_{{x \to -3^{+}}} f(x) = \frac{7}{0^+} = +\infty
$$

### c. $$ \lim_{{x \to -3^{-}}} f(x) $$

We evaluate the limit as $$ x $$ approaches $$-3$$ from the left (values less than $$-3$$).

$$
f(x) = \frac{4 - x}{3 + x}
$$

As $$ x \to -3^- $$:
- **Numerator:** $$ 4 - (-3) = 7 $$ (approaches 7)
- **Denominator:** $$ 3 + (-3^-) $$ approaches $$ 0^- $$ (a small negative number)

$$
\lim_{{x \to -3^{-}}} f(x) = \frac{7}{0^-} = -\infty
$$

### d. $$ \lim_{{x \to \infty}} f(x) $$

For the limit as $$ x $$ approaches infinity, we analyze the leading terms.

$$
f(x) = \frac{4 - x}{3 + x} = \frac{-x + 4}{x + 3}
$$

Divide numerator and denominator by $$ x $$:

$$
f(x) = \frac{-1 + \frac{4}{x}}{1 + \frac{3}{x}}
$$

As $$ x \to \infty $$:
- $$ \frac{4}{x} \to 0 $$
- $$ \frac{3}{x} \to 0 $$

$$
\lim_{{x \to \infty}} f(x) = \frac{-1 + 0}{1 + 0} = -1
$$

**Final Answer:**

**a.** The graph has a vertical asymptote at $$ x = -3 $$ and a horizontal asymptote at $$ y = -1 $$.

**b.** $$ \displaystyle \lim_{x \to -3^{+}} f(x) = +\infty $$

**c.** $$ \displaystyle \lim_{x \to -3^{-}} f(x) = -\infty $$

**d.** $$ \displaystyle \lim_{x \to \infty} f(x) = -1 $$
**a.** Vertical asymptote at $$ x = -3 $$ and horizontal asymptote at $$ y = -1 $$. **b.** As $$ x $$ approaches $$-3$$ from the right, $$ f(x) $$ approaches $$ +\infty $$. **c.** As $$ x $$ approaches $$-3$$ from the left, $$ f(x) $$ approaches $$ -\infty $$. **d.** As $$ x $$ approaches infinity, $$ f(x) $$ approaches $$-1$$.

Consider .
a. What are the asymptotes?
b. What is the ?
c. What is the ?
d. What is the ?

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