Q222 / 270 · NDA 2026 (Phase 1) · Mathematics
What is $\lim\limits_{x\to1}\dfrac{x^{(n^2-1)}-1}{x^{(n+1)}-1}$ equal to, where $n>1$ is a natural number ?
- A$0$
- B$1$
- Correct answerC$n-1$
- D$n+1$
Solution
The given limit is $\lim\limits_{x\to1}\dfrac{x^{(n^2-1)}-1}{x^{(n+1)}-1}$.
As $x \to 1$, the expression takes the indeterminate form $\dfrac{0}{0}$.
Applying L'Hospital's rule, we differentiate the numerator and the denominator with respect to $x$:
$\lim\limits_{x\to1}\dfrac{(n^2-1)x^{n^2-2}}{(n+1)x^n}$
Substituting $x=1$, we get:
$\dfrac{n^2-1}{n+1}$
$\Rightarrow \dfrac{(n-1)(n+1)}{n+1}$
$\Rightarrow n-1$
Answer: $n-1$
As $x \to 1$, the expression takes the indeterminate form $\dfrac{0}{0}$.
Applying L'Hospital's rule, we differentiate the numerator and the denominator with respect to $x$:
$\lim\limits_{x\to1}\dfrac{(n^2-1)x^{n^2-2}}{(n+1)x^n}$
Substituting $x=1$, we get:
$\dfrac{n^2-1}{n+1}$
$\Rightarrow \dfrac{(n-1)(n+1)}{n+1}$
$\Rightarrow n-1$
Answer: $n-1$
Hence, the correct answer is option: 'C'.