What is lim(x→1) [x^(n²-1) - 1] / [x^(n+1) - 1] equal to, where n > 1 is a natural number?
Answer
The correct answer is C: n - 1. Using the standard limit lim(x→1) (x^m - 1)/(x - 1) = m, we have lim(x→1) [x^(n²-1) - 1]/[x^(n+1) - 1] = (n² - 1)/(n + 1) = (n-1)(n+1)/(n+1) = n - 1.
A.0
B.1
C.n - 1✓ Correct
D.n + 1
Explanation
Using the standard limit lim(x→1) (x^m - 1)/(x - 1) = m, we have lim(x→1) [x^(n²-1) - 1]/[x^(n+1) - 1] = (n² - 1)/(n + 1) = (n-1)(n+1)/(n+1) = n - 1.
💡 Practice unlimited NDA PYQs + AI-tracked progress on each topic. Sign up free →