We are tasked with evaluating the limit:

$\lim_{x \to 9^+} \left( \frac{x - 9}{5 + \log_{\frac{1}{5}}(x - 9)} \right)$

Step 1: Behavior of $x - 9$ as $x \to 9^+$

As $x \to 9^+$, the expression $x - 9$ becomes a small positive number approaching $0^+$.

Step 2: Behavior of $\log_{\frac{1}{5}}(x - 9)$ as $x \to 9^+$

The logarithmic expression $\log_{\frac{1}{5}}(x - 9)$ involves a logarithm with a base $\frac{1}{5}$. Recall that logarithms with bases less than 1 have the following behavior:

$\log_{\frac{1}{5}}(x - 9) = \frac{\log(x - 9)}{\log(\frac{1}{5})}$

Since $\log(\frac{1}{5})$ is negative and $\log(x - 9)$ becomes negative as $x - 9 \to 0^+$, the entire logarithmic expression tends to $+\infty$ as $x - 9 \to 0^+$.

Step 3: Limit of the full expression

Now, we examine the full expression:

$\frac{x - 9}{5 + \log_{\frac{1}{5}}(x - 9)}$

As $x - 9 \to 0^+$:

• The numerator $x - 9 \to 0^+$.
• The denominator $5 + \log_{\frac{1}{5}}(x - 9) \to +\infty$.

Since the numerator approaches $0^+$ and the denominator approaches $+\infty$, the overall expression tends to $0$.

Conclusion:

The limit is:

$\lim_{x \to 9^+} \left( \frac{x - 9}{5 + \log_{\frac{1}{5}}(x - 9)} \right) = 0$

From the given options, the correct answer is:

$\boxed{E \, (0^-)}$

This is because the expression approaches $0$ from the negative side (due to the logarithmic behavior).

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Would you like further details on any part of this explanation?

Here are 5 questions to extend this topic:

1. How do logarithms behave for bases greater than 1 versus less than 1?
2. Can you provide other examples of limits involving logarithmic functions?
3. What are the key differences between one-sided and two-sided limits?
4. How would this limit change if the base of the logarithm were greater than 1?
5. Can you apply L'Hopital's Rule to verify the result?

Tip: For limits involving logarithmic functions, always analyze the behavior of the logarithm based on its base and the value inside the log.