Question

If \(\lim _{x \rightarrow 0} \frac{x^{2 n} \sin ^{n} x}{x^{2 n}-\sin ^{2 n} x}\) is a non zero finite number, then n must be equal to (A) 1 (B) 2 (C) 3 (D) none of these

Short answer

Answer: (D) none of these

Step-by-step solution

Apply L'Hopital's Rule

We need to differentiate the numerator and the denominator with respect to x. So, $$ \frac{d}{d x}\left(x^{2 n} \sin ^{n} x\right) = 2nx^{2n-1}\sin^n{x} + nx^{2n}\sin^{n-1}{x} \cos{x} $$ and $$ \frac{d}{d x}\left(x^{2 n}-\sin ^{2 n} x\right) = 2nx^{2n-1} - 2n\sin^{2n-1}{x} \cos{x} $$ Since we need the limit as x approaches 0, let's put x = 0 in the derivatives: $$ \lim _{x \rightarrow 0} \frac{d\left(x^{2 n} \sin ^{n} x\right)/dx}{d\left(x^{2 n}-\sin ^{2 n} x\right)/dx} = \lim _{x \rightarrow 0} \frac{ 2nx^{2n-1}\sin^n{x} + nx^{2n}\sin^{n-1}{x} \cos{x}}{2nx^{2n-1} - 2n\sin^{2n-1}{x} \cos{x}} $$

Simplify the Limit Expression

To simplify the limit, let's factor out the common terms: $$ \lim _{x \rightarrow 0} \frac{n(x^{2n}(1+\sin^{2n-1}{x}\cos{x})}{2n(x^{2n-1}(1-\sin^{2n-1}{x}\cos{x})} $$ Now, cancel the common terms of n and x^(2n-1) to get: $$ \lim _{x \rightarrow 0} \frac{1 + \sin^{2n-1}{x}\cos{x}}{2(1 - \sin^{2n-1}{x}\cos{x})} = \frac{1}{2} $$ Since the limit is a non-zero finite number, our analysis shows that: n must be equal to (D) none of these.