Let’s evaluate the given limits step-by-step and specify whether L’Hôpital’s Rule is applied.

(a) $\lim_{x \to \infty} \frac{2x^2 + 3x - 7}{2 + 3x - 7x^2}$

1. Reorganize terms by dividing numerator and denominator by $x^2$, the highest power in the denominator:

   = \frac{2 + \frac{3}{x} - \frac{7}{x^2}}{\frac{2}{x^2} + \frac{3}{x} - 7}.$$

2. Simplify as $x \to \infty$:

   • Terms involving $\frac{1}{x}$ and $\frac{1}{x^2}$ approach zero.
   • The result becomes:

   $\lim_{x \to \infty} \frac{2x^2 + 3x - 7}{2 + 3x - 7x^2} = \frac{2}{-7} = -\frac{2}{7}.$

3. No need for L’Hôpital’s Rule because the limit was evaluated by simplifying directly.

(b) $\lim_{x \to \infty} \frac{2x + 3}{e^{3x + 2}}$

1. Observe the growth rates:

   • The numerator ($2x + 3$) grows linearly.
   • The denominator ($e^{3x+2}$) grows exponentially. Exponential growth dominates.

2. Apply L’Hôpital’s Rule:

   • Rewrite as $\frac{f(x)}{g(x)}$, where $f(x) = 2x + 3$ and $g(x) = e^{3x+2}$.
   • Differentiate numerator and denominator:

   $f'(x) = 2, \quad g'(x) = 3e^{3x+2}.$

   The new fraction is:

   $\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{2}{3e^{3x+2}}.$

3. Simplify the limit:

   • As $x \to \infty$, $e^{3x+2} \to \infty$, so the fraction approaches:

   $\lim_{x \to \infty} \frac{2}{3e^{3x+2}} = 0.$

4. L’Hôpital’s Rule was used here.

(c) $\lim_{x \to \infty} \frac{\sqrt{2x + 3}}{\ln(3x + 2)}$

1. Determine growth rates:

   • The numerator ($\sqrt{2x+3}$) grows like $\sqrt{x}$.
   • The denominator ($\ln(3x+2)$) grows logarithmically. Square roots grow faster than logarithms as $x \to \infty$.

2. Direct simplification or substitution:

   • Divide both numerator and denominator by $\sqrt{x}$ for clarity.
   • Alternatively, rewrite the numerator as $(2x+3)^{1/2}$ and approximate growth as $x \to \infty$:

   $\sqrt{2x+3} \sim \sqrt{2x}, \quad \ln(3x+2) \sim \ln(x).$

   The limit becomes proportional to:

   $\frac{\sqrt{2x}}{\ln(x)}.$

3. Conclude: Since $\sqrt{x} \gg \ln(x)$, the limit grows without bound:

   $\lim_{x \to \infty} \frac{\sqrt{2x+3}}{\ln(3x+2)} = \infty.$

4. L’Hôpital’s Rule is not required here.

Final Answers:

• (a): $-\frac{2}{7}$, no L’Hôpital’s Rule.
• (b): $0$, using L’Hôpital’s Rule.
• (c): $\infty$, no L’Hôpital’s Rule.

Do you have any questions or need further clarification?

Related Questions:

1. What is the condition for using L’Hôpital’s Rule?
2. Can exponential functions always dominate polynomial growth in limits?
3. Why does $\ln(x)$ grow slower than $\sqrt{x}$?
4. What happens if the numerator and denominator grow at the same rate in a limit?
5. How does dividing by the highest power of $x$ simplify rational expressions?

Tip:

When comparing growth rates, remember this order: exponential $>$ polynomial $>$ logarithmic, as $x \to \infty$.