calculate the lim as n goes to 9 of : (x^2-81)/(6ln(x+8)) using series

To calculate the limit as \( n \) approaches 9 of the function \( \frac{x^2 - 81}{6\ln(x + 8)} \) using series expansion, we proceed as follows:

### Step 1: Rewrite the function

Notice that when \( x = 9 \), the numerator \( x^2 - 81 \) becomes \( 9^2 - 81 = 0 \), and the denominator \( 6\ln(x + 8) \) becomes \( 6\ln(9 + 8) = 6\ln(17) \). Therefore, we have a \( \frac{0}{0} \) indeterminate form, which suggests that we can apply a series expansion to simplify the problem.

### Step 2: Series expansion around \( x = 9 \)

#### Numerator Expansion:
The numerator is \( x^2 - 81 \). We can factor this as:
\[
x^2 - 81 = (x - 9)(x + 9)
\]
Expanding around \( x = 9 \):
\[
x^2 - 81 = (x - 9)(x + 9) = (x - 9)(9 + (x - 9)) = (x - 9)(18 + (x - 9))
\]

This simplifies to:
\[
x^2 - 81 = 18(x - 9) + (x - 9)^2
\]

#### Denominator Expansion:
The denominator is \( 6\ln(x + 8) \). We can expand \( \ln(x + 8) \) around \( x = 9 \) using the Taylor series expansion for \( \ln(1 + y) \) where \( y = \frac{x - 9}{17} \) (since \( x + 8 = 17 + (x - 9) \)):
\[
\ln(1 + y) \approx y - \frac{y^2}{2} + \dots
\]

So:
\[
\ln(x + 8) = \ln(17 + (x - 9)) = \ln(17) + \frac{x - 9}{17} - \frac{(x - 9)^2}{2 \times 17^2} + \dots
\]

Multiplying by 6:
\[
6\ln(x + 8) = 6\ln(17) + \frac{6(x - 9)}{17} - \frac{3(x - 9)^2}{17^2} + \dots
\]

### Step 3: Simplify the limit

Now, substitute these series expansions into the original expression:
\[
\lim_{x \to 9} \frac{18(x - 9) + (x - 9)^2}{6\ln(17) + \frac{6(x - 9)}{17} - \frac{3(x - 9)^2}{17^2} + \dots}
\]

The dominant term in the numerator and denominator as \( x \) approaches 9 is the linear term \( (x - 9) \), so we can write:
\[
\lim_{x \to 9} \frac{18(x - 9)}{\frac{6(x - 9)}{17}} = \lim_{x \to 9} \frac{18 \cdot 17}{6} = \frac{18 \cdot 17}{6} = 51
\]

Thus, the limit is \( 51 \).

### Final Answer:
\[
\lim_{x \to 9} \frac{x^2 - 81}{6\ln(x + 8)} = 51
\]

Would you like further details, or have any questions about this solution?

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1. How do you determine if a series expansion is needed for a limit?
2. What is the significance of a \( \frac{0}{0} \) indeterminate form?
3. How is the Taylor series used to approximate logarithmic functions?
4. Can you apply this method to other limits involving indeterminate forms?
5. What happens if the dominant term cancels out in both the numerator and the denominator?
6. How do higher-order terms affect the limit calculation?
7. How does the factorization of the numerator help in simplifying the limit?
8. What are some common pitfalls when using series expansion for limits?

**Tip:** When using series expansion to evaluate limits, always check the order of the terms in the expansion to determine which term dominates as the variable approaches the limit point.

Learn how to calculate the limit as x approaches 9 of (x^2 - 81) divided by 6 times the natural logarithm of (x + 8) using series expansion. This detailed solution involves series expansion around x = 9, Taylor series for logarithmic functions, and handling indeterminate forms. Suitable for advanced mathematics students.

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