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:

L'Hôpital's Rule is often the fastest way to evaluate \(\frac{0}{0}\) limits of polynomials:
Differentiate the numerator and denominator with respect to \(x\):
\[ \lim_{x \to -a} \frac{\frac{d}{dx}(x^9 + a^9)}{\frac{d}{dx}(x + a)} = \lim_{x \to -a} \frac{9x^8}{1} = 9(-a)^8 = 9a^8 \] Then solve \(9a^8 = 9 \implies a = \pm 1\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a limit equation: \(\lim_{x \to -a} \frac{x^9 + a^9}{x + a} = 9\).
We need to find the possible real values of \(a\).
Key Formula or Approach:
Direct substitution of \(x = -a\) gives \(\frac{(-a)^9 + a^9}{-a + a} = \frac{0}{0}\) (since 9 is an odd power), which is an indeterminate form.
We can use L'Hôpital's Rule or the standard algebraic limit formula:
\[ \lim_{x \to u} \frac{x^n - u^n}{x - u} = n u^{n-1} \]

Step 2: Detailed Explanation:

• Let us rewrite the limit expression to match the standard formula:
Let \(u = -a\). Then \(a = -u\) and \(a^9 = (-u)^9 = -u^9\).
The limit becomes:
\[ \lim_{x \to u} \frac{x^9 - u^9}{x - u} \]

• Using the limit formula \(\lim_{x \to u} \frac{x^n - u^n}{x-u} = n u^{n-1}\) with \(n = 9\):
\[ \lim_{x \to u} \frac{x^9 - u^9}{x - u} = 9 u^{9-1} = 9 u^8 \]

• Substitute \(u = -a\) back into the expression:
\[ 9 (-a)^8 = 9 a^8 \]

• Set this equal to the given limit value of 9:
\[ 9 a^8 = 9 \] \[ a^8 = 1 \]

• Solving for real values of \(a\):
Since \(a \in \mathbb{R}\), the equation \(a^8 = 1\) has only two real roots:
\[ a = \pm 1 \]

Step 3: Final Answer:
The possible values of \(a\) are \(\pm 1\).