Evaluating $\lim_{x\to 0}\dfrac{\cos(\frac{\pi}{2}\cos x)}{\sin(\sin x)}$ without L'Hôpital's rule

Question

Without using L'Hôpital's rule, find: $$\lim_{x\to 0}\dfrac{\cos(\frac{\pi}{2}\cos x)}{\sin(\sin x)}$$ I know that the answer is $0$.

My attempt:

I tried by using the half-angle formula, $$\cos\left(\frac{A}{2}\right)=\pm\sqrt{\frac{1+\cos A}{2}},$$ but I've couldn't find it.

Answer 1

\begin{aligned}\lim_{x\to 0}{\frac{\cos{\left(\frac{\pi}{2}\cos{x}\right)}}{\sin{\left(\sin{x}\right)}}}&=\lim_{x\to 0}{\frac{\sin{\left(\frac{\pi}{2}-\frac{\pi}{2}\cos{x}\right)}}{\sin{\left(\sin{x}\right)}}}\\ &=\lim_{x\to 0}{\left(\frac{\pi x}{2}\times\frac{1-\cos{x}}{x^{2}}\times\frac{\sin{\left(\frac{\pi}{2}-\frac{\pi}{2}\cos{x}\right)}}{\frac{\pi}{2}-\frac{\pi}{2}\cos{x}}\times\frac{x}{\sin{x}}\times\frac{\sin{x}}{\sin{\left(\sin{x}\right)}}\right)}\\ &=0\times\frac{1}{2}\times 1\times 1\times 1\\ &=0\end{aligned}

Because $$ \lim_{\textrm{stuff}\to 0}{\frac{\sin{\left(\mathrm{stuff}\right)}}{\mathrm{stuff}}}=\lim_{x\to 0}{\frac{\sin{x}}{x}}=1 $$

And $$ \lim_{x\to 0}{\frac{1-\cos{x}}{x^{2}}}=\lim_{x\to 0}{\frac{1}{2}\left(\frac{\sin{\left(\frac{x}{2}\right)}}{\frac{x}{2}}\right)^{2}}=\frac{1}{2} $$

Answer 2

Rewrite the limit as a product like so:

$$L = \lim_{x\to 0} \frac{\sin x}{\sin( \sin x)} \cdot\frac{x}{\sin x} \cdot \frac{\cos(\frac{\pi}{2}\cos x)}{x}$$

If the limits exist individually then the limit of their product will be the product of their limits. The first two limits are $1$, giving us that

$$L = \lim_{x\to 0} \frac{\cos(\frac{\pi}{2}\cos x)}{x} = \lim_{x\to 0} \frac{\cos(\frac{\pi}{2}\cos x) - \cos(\frac{\pi}{2}\cos 0)}{x-0} \equiv \Bigr(\cos(\frac{\pi}{2}\cos x)\Bigr)'\Bigr|_{x=0} = 0$$

by the definition of the derivative.

Answer 3

If you want more than the limit itself, compose Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cos(\frac{\pi}{2}\cos (x))=\frac{\pi x^2}{4}-\frac{\pi x^4}{48}+O\left(x^6\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin(\sin(x))=x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^6\right)$$ $$\frac{\cos(\frac{\pi}{2}\cos (x)) } {\sin(\sin(x)) }=\frac{\frac{\pi x^2}{4}-\frac{\pi x^4}{48}+O\left(x^6\right) } {x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^6\right) }=\frac{\pi x}{4}+\frac{\pi x^3}{16}+O\left(x^5\right)$$