Let's analyze the two limit problems provided in the image.

Problem (a):

$\lim_{x \to \frac{7}{4}^+} \frac{-14x}{7 - 4x}$ Here, we need to evaluate the limit as $x$ approaches $\frac{7}{4}$ from the right ($x \to \frac{7}{4}^+$).

1. Substitute $x = \frac{7}{4}$ into the denominator: $7 - 4x = 7 - 4 \times \frac{7}{4} = 7 - 7 = 0$ Since the denominator approaches 0, we need to check the sign of the denominator as $x$ approaches $\frac{7}{4}$ from the right.

   • If $x > \frac{7}{4}$, then $7 - 4x < 0$, meaning the denominator is negative.

2. Analyze the numerator: $-14x \quad \text{as} \quad x \to \frac{7}{4}^+$ The numerator $-14x$ is negative as well, since $-14 \times \frac{7}{4} = -24.5$.

3. Behavior of the limit: As $x$ approaches $\frac{7}{4}$ from the right, the numerator approaches a negative number, and the denominator approaches a small negative value (since $7 - 4x$ becomes slightly negative). The overall fraction thus approaches positive infinity because the fraction is negative divided by negative: $\lim_{x \to \frac{7}{4}^+} \frac{-14x}{7 - 4x} = +\infty$

Problem (b):

$\lim_{x \to \frac{7}{4}^-} \frac{-14x}{7 - 4x}$ This time, we are evaluating the limit as $x$ approaches $\frac{7}{4}$ from the left ($x \to \frac{7}{4}^-$).

1. Substitute $x = \frac{7}{4}$ into the denominator (same as before): $7 - 4x = 0$ As $x$ approaches $\frac{7}{4}$ from the left, $7 - 4x > 0$, meaning the denominator is positive.

2. Analyze the numerator: Again, the numerator $-14x$ is negative.

3. Behavior of the limit: As $x$ approaches $\frac{7}{4}$ from the left, the numerator is negative, and the denominator approaches a small positive value. This results in the fraction approaching negative infinity: $\lim_{x \to \frac{7}{4}^-} \frac{-14x}{7 - 4x} = -\infty$

Final Answers:

• (a) $\lim_{x \to \frac{7}{4}^+} \frac{-14x}{7 - 4x} = +\infty$
• (b) $\lim_{x \to \frac{7}{4}^-} \frac{-14x}{7 - 4x} = -\infty$

Would you like further details or have any questions?

Here are five related questions to further your understanding:

1. How does the behavior of the denominator influence the sign of the limit?
2. Why do we need to check the direction from which $x$ approaches $\frac{7}{4}$?
3. Can you find another limit where the numerator and denominator both approach zero?
4. How would the limit change if the function were $\frac{14x}{7 - 4x}$ instead of $\frac{-14x}{7 - 4x}$?
5. What happens to the limit if the numerator is constant instead of depending on $x$?

Tip: When solving limits involving fractions, always check the signs of both the numerator and denominator as $x$ approaches the given value.