Question

If $\lim_{x\to 0} \frac{\sin x \sin 2x \sin 3x...\sin nx - x^n}{x \tan((1+x)(1+2x)(1+3x)....(1+2023x))}$ exists and is non-zero, then 'n' equals

• A. 2022
• B. 2023
• C. 2024
• D. no such value of n exists

Answer 1

Answer: (D)

no such value of n exists

Answer 2

Let the limit be $L$. The numerator is $N(x) = \prod_{k=1}^{n} \sin kx - x^n$ and the denominator is $D(x) = x \tan(\prod_{k=1}^{2023} (1+kx))$.

For the denominator, as $x \to 0$, $\prod_{k=1}^{2023} (1+kx) \to 1$. Thus, $\tan(\prod_{k=1}^{2023} (1+kx)) \to \tan(1)$. The denominator behaves as $D(x) \sim x \tan(1)$ for $x \to 0$. The lowest power of $x$ is $x^1$.

For the numerator, using the Taylor expansion $\sin y = y - \frac{y^3}{6} + O(y^5)$: $\prod_{k=1}^{n} \sin kx = \prod_{k=1}^{n} (kx - \frac{(kx)^3}{6} + O(x^5))$ $= (x - \frac{x^3}{6} + ...)(2x - \frac{(2x)^3}{6} + ...)...(nx - \frac{(nx)^3}{6} + ...)$ The term with the lowest power of $x$ in the product is $(x)(2x)...(nx) = n! x^n$. So, $\prod_{k=1}^{n} \sin kx = n! x^n + O(x^{n+2})$.

The numerator is $N(x) = n! x^n + O(x^{n+2}) - x^n = (n!-1)x^n + O(x^{n+2})$.

For the limit $L$ to exist and be non-zero, the lowest power of $x$ in the numerator must match the lowest power of $x$ in the denominator, which is $x^1$.

Case 1: $n > 1$. The lowest power of $x$ in the numerator is $x^n$. For this to match the denominator's $x^1$, we need $n=1$. This contradicts the assumption $n>1$.

Case 2: $n = 1$. The numerator is $N(x) = \sin x - x$. Using Taylor expansion, $\sin x - x = (x - \frac{x^3}{6} + O(x^5)) - x = -\frac{x^3}{6} + O(x^5)$. The lowest power of $x$ in the numerator is $x^3$. The denominator's lowest power is $x^1$. The limit becomes $L = \lim_{x\to 0} \frac{-x^3/6}{x \tan(1)} = \lim_{x\to 0} \frac{-x^2}{6 \tan(1)} = 0$. This limit is zero, not non-zero.

Since neither $n=1$ nor $n>1$ yields a non-zero limit, there is no such value of $n$ that satisfies the given condition.