Question

If \(\lim_{x \to -a} \frac{x^9 + a^9}{x + a} = 9\), then the possible values of a are given by :

• A. \(a = \pm 1\)
• B. \(a = \pm 9\)
• C. \(a = \pm 3\)
• D. \(a = \pm 8\)

Hint:

With L'Hôpital's Rule, limits of the form \(\frac{x^n - u^n}{x - u}\) can be computed in a single step by simply taking the derivative of the numerator and evaluating it:
\[ \left. \frac{d}{dx}(x^n) \right|_{x=u} = n u^{n-1} \]
For this problem: \(9 (-a)^8 = 9 \implies a^8 = 1 \implies a = \pm 1\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To evaluate this limit, we can use a standard algebraic limit formula or apply L'Hôpital's Rule, since substituting \(x = -a\) directly into the expression yields an indeterminate form of \(\frac{0}{0}\).
Key Formula or Approach:
The standard algebraic limit formula is:
\[ \lim_{x \to u} \frac{x^n - u^n}{x - u} = n u^{n-1} \]
Alternatively, we can use L'Hôpital's Rule by differentiating the numerator and the denominator with respect to \(x\).

Step 2: Detailed Explanation:
Let us use L'Hôpital's Rule to find the limit.
The given limit is:
\[ \lim_{x \to -a} \frac{x^9 + a^9}{x + a} \]
As \(x \to -a\), the numerator approaches \((-a)^9 + a^9 = -a^9 + a^9 = 0\) (since 9 is an odd exponent).
The denominator approaches \((-a) + a = 0\).
Since it is of the indeterminate form \(\frac{0}{0}\), we differentiate the numerator and denominator with respect to \(x\):
- Derivative of numerator: \(\frac{d}{dx}(x^9 + a^9) = 9x^8\)
- Derivative of denominator: \(\frac{d}{dx}(x + a) = 1\)
Now, calculate the limit:
\[ \lim_{x \to -a} \frac{9x^8}{1} = 9(-a)^8 = 9a^8 \]
We are given that this limit is equal to 9:
\[ 9a^8 = 9 \]
Divide both sides by 9:
\[ a^8 = 1 \]
Taking the 8th root on both sides for real values of \(a\) gives:
\[ a = \pm 1 \]
Thus, the possible values of \(a\) are \(\pm 1\). This matches Option (A).

Step 3: Final Answer:
The correct option is (A).