Question

The value of $\lim_{x \to -1} \frac{x^{10} - 1}{x + 1}$ is given by}

• A. 10
• B. 0
• C. $\infty$
• D. -10

Hint:

Alternatively, use the standard algebraic limit formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.
Here, let $a = -1$, $n = 10$:
\[ \lim_{x \to -1} \frac{x^{10} - (-1)^{10}}{x - (-1)} = 10 \cdot (-1)^9 = -10 \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Direct substitution of $x = -1$ into the limit yields the indeterminate form $0/0$.
We can resolve this indeterminate form using L'Hôpital's Rule or algebraic factorization.
Key Formula or Approach:
L'Hôpital's Rule states that if $\lim \frac{f(x)}{g(x)} = \frac{0}{0}$, then:
\[ \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} \]

Step 2: Detailed Explanation:
We want to evaluate:
\[ L = \lim_{x \to -1} \frac{x^{10} - 1}{x + 1} \]
Check direct substitution of $x = -1$:
- Numerator: $(-1)^{10} - 1 = 1 - 1 = 0$
- Denominator: $-1 + 1 = 0$
Since it is of the form $0/0$, apply L'Hôpital's Rule by differentiating the numerator and denominator with respect to $x$:
- Derivative of numerator: $\frac{d}{dx}(x^{10} - 1) = 10x^9$
- Derivative of denominator: $\frac{d}{dx}(x + 1) = 1$
Substitute these derivatives back into the limit:
\[ L = \lim_{x \to -1} \frac{10x^9}{1} \]
Now, evaluate the limit by direct substitution of $x = -1$:
\[ L = 10 \cdot (-1)^9 = 10 \cdot (-1) = -10 \]
Therefore, the value of the limit is $-10$.

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