Question

question 9 evaluate the following limits. consider looking at the graph of the function to help you understand the situation. as necessary, enter oo for ∞ and -oo for −∞. (a) \\( \lim_{x \to \frac{3}{8}^+} \left( \frac{38x}{3 - 8x} \
ight) = \\) (b) \\( \lim_{x \to \frac{3}{8}^-} \left( \frac{38x}{3 - 8x} \
ight) = \\)

Explanation:

Step1: Identify the vertical asymptote

First, find where the denominator is zero:
$3-8x=0 \implies x=\frac{3}{8}$

Step2: Analyze right-hand limit (a)

For $x \to \frac{3}{8}^+$, $x>\frac{3}{8}$, so $3-8x<0$. Substitute a value slightly larger than $\frac{3}{8}$, e.g., $x=0.38$:
$3-8(0.38)=3-3.04=-0.04$, numerator $38(0.38)=14.44$.
$\lim_{x \to \frac{3}{8}^+} \frac{38x}{3-8x} = \frac{\text{Positive}}{\text{Negative}} = -\infty$

Step3: Analyze left-hand limit (b)

For $x \to \frac{3}{8}^-$, $x<\frac{3}{8}$, so $3-8x>0$. Substitute a value slightly smaller than $\frac{3}{8}$, e.g., $x=0.37$:
$3-8(0.37)=3-2.96=0.04$, numerator $38(0.37)=14.06$.
$\lim_{x \to \frac{3}{8}^-} \frac{38x}{3-8x} = \frac{\text{Positive}}{\text{Positive}} = +\infty$

Answer:

(a) $-\infty$
(b) $\infty$