https://r2.erweima.ai/mathbot/image/1726834598494zagmky36.jpg

The expression given in the image is a limit problem. It is written as:

\[
\lim_{n \to \infty} \frac{(-2)^n + 5^n}{(-2)^{n+1} + 5^{n+1}}.
\]

To solve this limit as \( n \to \infty \), let's analyze the terms step by step.

### Step 1: Growth of the Terms
- The expression contains powers of \( (-2) \) and \( 5 \). As \( n \to \infty \), the term \( 5^n \) grows significantly faster than \( (-2)^n \). This is because \( 5 > 2 \), so \( 5^n \) dominates \( (-2)^n \) for large \( n \).

### Step 2: Factor out \( 5^n \)
To make the analysis clearer, factor \( 5^n \) from both the numerator and denominator:

\[
\frac{(-2)^n + 5^n}{(-2)^{n+1} + 5^{n+1}} = \frac{5^n \left(\frac{(-2)^n}{5^n} + 1 \right)}{5^{n+1} \left(\frac{(-2)^{n+1}}{5^{n+1}} + 1 \right)}.
\]

### Step 3: Simplify Each Term
- As \( n \to \infty \), \( \frac{(-2)^n}{5^n} \to 0 \) because \( \left(\frac{-2}{5}\right)^n \) becomes smaller and smaller.
- Similarly, \( \frac{(-2)^{n+1}}{5^{n+1}} \to 0 \) for large \( n \).

Thus, the expression simplifies to:

\[
\frac{5^n(1)}{5^{n+1}(1)} = \frac{1}{5}.
\]

### Final Answer:
\[
\lim_{n \to \infty} \frac{(-2)^n + 5^n}{(-2)^{n+1} + 5^{n+1}} = \frac{1}{5}.
\]

Would you like a more detailed explanation or have any further questions?

Here are some related questions for further practice:
1. How would the limit change if the base was \( 3 \) instead of \( 5 \)?
2. Can you generalize the limit behavior when you have two terms with different bases?
3. How does the behavior of negative powers affect the outcome of limits?
4. What happens to the limit if both bases were negative, say \( (-3) \) and \( (-4) \)?
5. How does this approach change for finite \( n \) values instead of \( n \to \infty \)?

**Tip:** When evaluating limits involving powers, always look for the fastest-growing term, as it will dominate the behavior of the entire expression.

Find the limit: lim (n -> infinity) ((-2)^n + 5^n) / ((-2)^(n+1) + 5^(n+1)).

Learn how to solve the limit lim (n -> infinity) ((-2)^n + 5^n) / ((-2)^(n+1) + 5^(n+1)), involving the asymptotic behavior of exponential terms. Explore how different powers affect the limit outcome. Suitable for advanced high school students in Grades 11-12.

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