Question

# Find $$n \in N$$. If $$\lim_{x \to 0}(1 – \cos x \cdot \cos 2x \cdot \cos 3x …. \cos nx) = 253$$

Solution:

Since given expression $$\to \frac{0}{0}$$ , We can use L’Hopital

$$\lim_{x \to 0} \frac{0 – [(-\sin x)(\cos 2x\cdot \cos 3x \cos 4x …. \cos nx) + (-2\sin 2x)(\cos x\cdot \cos 3x \cos 4x …. \cos nx) + (-3\sin 3x)(\cos x \cdot \cos 2x \cos 4x … \cos nx) ….. (-n \sin nx)(\cos x \cdot \cos 2x \cdot \cos 3x …. \cos (n-1))]}{2x}$$

$$\lim_{x \to 0} \frac{(\sin x)(\cos 2x\cdot \cos 3x \cos 4x …. \cos nx) + (2\sin 2x)(\cos x\cdot \cos 3x \cos 4x …. \cos nx) + (3\sin 3x)(\cos x \cdot \cos 2x \cos 4x … \cos nx) ….. (n \sin nx)(\cos x \cdot \cos 2x \cdot \cos 3x …. \cos (n-1))}{2x}$$

$$\frac{1}{2} \lim_{x \to 0} \frac{(\sin x)(\cos 2x\cdot \cos 3x \cos 4x …. \cos nx) + (2\sin 2x)(\cos x\cdot \cos 3x \cos 4x …. \cos nx) + (3\sin 3x)(\cos x \cdot \cos 2x \cos 4x … \cos nx) ….. (n \sin nx)(\cos x \cdot \cos 2x \cdot \cos 3x …. \cos (n-1))}{x}$$

$$\frac{1}{2} \lim_{x \to 0} \frac{(\sin x)(\cos 2x\cdot \cos 3x \cos 4x …. \cos nx)}{x} + \frac{2\cdot (2\sin 2x)(\cos x\cdot \cos 3x \cos 4x …. \cos nx)}{2x} + \frac{3\cdot (3\sin 3x)(\cos x \cdot \cos 2x \cos 4x … \cos nx)}{3x} ….. \frac{n\cdot (n \sin nx)(\cos x \cdot \cos 2x \cdot \cos 3x …. \cos (n-1))}{nx}$$

$$\frac{1}{2} \lim_{x \to 0} \frac{(\sin x)(\cos 2x\cdot \cos 3x \cos 4x …. \cos nx)}{x} + \frac{2^2 (\sin 2x)(\cos x\cdot \cos 3x \cos 4x …. \cos nx)}{2x} + \frac{3^2(\sin 3x)(\cos x \cdot \cos 2x \cos 4x … \cos nx)}{3x} ….. \frac{n^2( \sin nx)(\cos x \cdot \cos 2x \cdot \cos 3x …. \cos (n-1))}{nx}$$

we know $$\lim_{t \to 0}\cos t \to 0 \; and \; \lim_{t \to 0} \frac{\sin t}{t} \to 1$$

Therefore $$\frac{1}{2}[1 + 2^2 + 3^2 + …. + n^2]$$
$$= \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6}$$

From question $$\frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} = 253$$
$$\Rightarrow n(n+1)(2n+1) = 12 \cdot \ 11 \cdot 23$$

Hence $$n = 11$$

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