We are tasked with finding the limit of the following expression as $x$ approaches infinity:

$\lim_{x \to \infty} \frac{25^x - 1}{16^x - 1}$

Step-by-Step Solution:

1. Rewrite the expression: $\lim_{x \to \infty} \frac{25^x - 1}{16^x - 1}$ Since both $25^x$ and $16^x$ grow exponentially as $x \to \infty$, the terms "-1" in both the numerator and the denominator become negligible. Thus, we approximate the limit by simplifying the expression as: $\lim_{x \to \infty} \frac{25^x}{16^x}$

2. Simplify the fraction: $\frac{25^x}{16^x} = \left( \frac{25}{16} \right)^x$ Now we need to evaluate the limit of $\left( \frac{25}{16} \right)^x$ as $x \to \infty$.

3. Evaluate the base: The base of the exponent is $\frac{25}{16} = 1.5625$, which is greater than 1. When a number greater than 1 is raised to a power that approaches infinity, the result also tends towards infinity.

   Therefore: $\lim_{x \to \infty} \left( \frac{25}{16} \right)^x = \infty$

Conclusion:

$\lim_{x \to \infty} \frac{25^x - 1}{16^x - 1} = \infty$

As $x$ approaches infinity, the expression grows without bound.

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Do you want more details or have any further questions?

Here are 5 related questions:

1. How does the behavior of exponential functions differ when the base is less than 1?
2. What is L'Hôpital's rule and how could it apply to limits involving infinity?
3. Can you describe the difference between polynomial and exponential growth?
4. What happens to the expression if the bases were reversed (e.g., $16^x$ in the numerator)?
5. How would logarithms simplify the analysis of similar limits?

Tip: When evaluating limits, simplifying the expression early can help avoid unnecessary complexity.