NDA 2026 — Mathematics PYQ

The given limit is:

$$\underset{x\to 1}{\lim }\frac{x^{(n^2-1)}-1}{x^{(n+1)}-1}$$

If we substitute $x=1$ directly, we get the indeterminate form $\frac{0}{0}$. Therefore, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately with respect to $x$.

1. Differentiating the numerator and denominator:

$$\underset{x\to 1}{\lim }\frac{\frac{d}{dx}(x^{(n^2-1)}-1)}{\frac{d}{dx}(x^{(n+1)}-1)}$$

2. Applying the power rule $\frac{d}{dx}(x^k)=k{x}^{k-1}$:

$$\underset{x\to 1}{\lim }\frac{(n^2-1)x^{(n^2-1-1)}}{(n+1)x^{(n+1-1)}}$$

$$\underset{x\to 1}{\lim }\frac{(n^2-1)x^{n^2-2}}{(n+1)x^n}$$

3. Substituting $x=1$:

$$\frac{(n^2-1)(1{)}^{n^2-2}}{(n+1)(1{)}^n}=\frac{n^2-1}{n+1}$$

4. Simplifying the expression:

Using the algebraic identity $a^2-b^2=(a-b)(a+b)$, we can factor $n^2-1$:

$$\frac{(n-1)(n+1)}{(n+1)}$$

$$=n-1$$

Final Answer:

The correct option is (c) $n-1$.