Let's analyze and solve the given limit using L'Hopital's Rule step by step.

The problem is:

$\lim_{x \to \infty} \frac{\ln(2x) + 5x^2}{2^x - 1}.$

Step 1: Check for the indeterminate form

• As $x \to \infty$:
  • $\ln(2x) + 5x^2 \to \infty$ because $5x^2 \to \infty$.
  • $2^x - 1 \to \infty$ because $2^x \to \infty$.
• Thus, the limit is in the form $\frac{\infty}{\infty}$, which allows the application of L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule

Differentiate the numerator and denominator with respect to $x$:

1. Numerator:
   $\frac{d}{dx} \left( \ln(2x) + 5x^2 \right) = \frac{1}{2x} \cdot 2 + 10x = \frac{1}{x} + 10x.$

2. Denominator:
   $\frac{d}{dx} \left( 2^x - 1 \right) = \ln(2) \cdot 2^x.$

Thus, the new limit becomes: $\lim_{x \to \infty} \frac{\frac{1}{x} + 10x}{\ln(2) \cdot 2^x}.$

Step 3: Simplify and check indeterminate form

As $x \to \infty$:

• The term $\frac{1}{x} \to 0$, so the numerator simplifies to $10x$.
• The denominator grows exponentially as $\ln(2) \cdot 2^x$.

The new limit is still in the form $\frac{\infty}{\infty}$, so we apply L'Hopital's Rule again.

Step 4: Apply L'Hopital's Rule again

Differentiate the numerator and denominator again:

1. Numerator:
   $\frac{d}{dx} \left( 10x \right) = 10.$

2. Denominator:
   $\frac{d}{dx} \left( \ln(2) \cdot 2^x \right) = \ln(2) \cdot \ln(2) \cdot 2^x = (\ln(2))^2 \cdot 2^x.$

Thus, the new limit becomes: $\lim_{x \to \infty} \frac{10}{(\ln(2))^2 \cdot 2^x}.$

Step 5: Evaluate the final limit

• As $x \to \infty$, the denominator $(\ln(2))^2 \cdot 2^x \to \infty$.
• Thus, the entire fraction approaches: $\frac{10}{\infty} = 0.$

Final Answer:

[ \lim_{x \to \infty} \frac{\ln(2x) + 5x