Question

The value of $ \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} $ is:

• A. \( \frac{2}{3} \)
• B. 1
• C. \( \frac{3}{2} \)
• D. Does not exist

Hint:

When you encounter indeterminate forms like \( \frac{0}{0} \) in limits, apply L'Hopital's Rule by taking the derivative of the numerator and denominator, then evaluate the limit.

Answer 1

The Correct Option is C

We are given the limit: \[ \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} \] We can apply L'Hopital's Rule to evaluate the limit, as this is an indeterminate form \( \frac{0}{0} \) when \( x = 1 \). L'Hopital's Rule states that if the limit is in the form \( \frac{0}{0} \), then: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] First, compute the derivatives of the numerator and denominator: The derivative of the numerator \( f(x) = x^{15} - 1 \) is: \[ f'(x) = 15x^{14} \] The derivative of the denominator \( g(x) = x^{10} - 1 \) is: \[ g'(x) = 10x^9 \] Now, apply L'Hopital's Rule: \[ \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} = \lim_{x \to 1} \frac{15x^{14}}{10x^9} \] Evaluating the limit at \( x = 1 \): \[ = \frac{15(1)^{14}}{10(1)^9} = \frac{15}{10} = \frac{3}{2} \] Thus, the value of the limit is: \[ \frac{3}{2} \]