Functions Period

Algebra Level 4

If period of s i n 2 m ( k x ) , m ∈ N sin^{2m}(\sqrt{k}x),m\in \mathbb N is π \pi , then lim ⁡ n → ∞ k n = \displaystyle \lim_{n\to \infty}k^{n}=

Also see this problem


The answer is 1.00.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

The period of all sin ⁡ 2 m ( x ) , m ∈ N \sin^{2m}(x), m\in\mathbb{N} funtions is π \pi

That would mean k x = x \sqrt{k}x = x

k = 1 \sqrt{k}=1

lim ⁡ n → ∞ k n = lim ⁡ n → ∞ 1 n = 1 \lim_{n \to \infty}k^{n}=\lim_{n \to \infty}1^{n}=1

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