How do I find the limit of $\lim_{x \to 0}{\frac{1-(\cos{kx})^k}{x^2}}$?

Question

As the title suggests, how do I find: $\;\;\displaystyle\lim_{x \to 0}{\frac{1-(\cos{kx})^k}{x^2}}?$

Our professor showed us how to find $\lim_{x \to 0}{\frac{1-\cos{kx}}{x^2}}$ by applying $1-\cos{x} = 2(\sin{\frac{x}{2}})^2$, but I have no idea how to even start this one.

We're not supposed to use derivatives, as we haven't reached that point or anything higher (obviously).

Answer 1

We can use Taylor expansion of the cosine function near the origin: $$\cos(kx)=1-\frac{k^2}{2}x^2+o(x^2)$$

We can derive this formula, also, looking at the asymptotic relations: $$\cos(kx)\;\;\sim\;\; 1-\frac{k^2}{2}x^2$$ In this case, it's not necessary to use the $o(x^2)$. I will keep using it because it's more formal.

So: $$1-\cos(kx)^k=1-\left(1-\frac{k^2}{2}x^2+o(x^2)\right)^k=\frac{k^3}{2}x^{2}+o(x^2)$$

The limit becomes: $$\lim_{x \to 0}{\frac{1-(\cos{kx})^k}{x^2}}\;\;\sim\;\; \lim_{x\to 0}\frac{\frac{k^3}{2}x^2+o(x^2)}{x^2}=\frac{1}{2}k^3$$

Question for you: what about the limit $$\lim_{x \to 0}{\frac{1-(\cos({\alpha x}))^{\beta}}{x^2}}$$

Answer 2

Notice that :

1. For $k = 2$ : $$\dfrac{1 - \cos^2(2x)}{x^2} = \dfrac{1 - \cos(2x)}{x^2} + \cos (2 x) \dfrac{1 - \cos(2x)}{x^2}$$
2. For $k = 3$ : $$\dfrac{1 - \cos^3(3x)}{x^2} = \dfrac{1 - \cos(3x)}{x^2} + \cos (3 x) \dfrac{1 - \cos(3x)}{x^2} + \cos^2(3x) \dfrac{1 - \cos(3x)}{x^2}$$
3. In general : $$\dfrac{1 - \cos^k(kx)}{x^2} = \dfrac{1 - \cos(kx)}{x^2} + \cos (kx) \dfrac{1 - \cos(kx)}{x^2} + \cdots + \cos^{k - 1}(kx) \dfrac{1 - \cos(kx)}{x^2}$$