So I am having trouble finding this limit: $$\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x}$$
The problem is I can't use the derivative of the composition of two functions nor can I use other techniques like l'Hôpital's theorem. I tried numerous techniques to calculate this limit but in vain so if you have any simple idea that is in the scope of my knowledge ( I am a pre-calculus student ), please do let me know without actually answering the question.
On
$$sin(\alpha) = sin(\pi - \alpha)$$ $$sin(\pi\sqrt{cos{x}}) = sin(\pi -\pi\sqrt{cos{x}})$$ $$ sin(\pi -\pi\sqrt{cos{x}}) \sim_{x\to 0} \pi -\pi\sqrt{cos{x}}$$ $$\lim_{x\to 0} \frac{sin(\pi\sqrt{cos{x}})}{x} = \lim_{x\to 0} \frac{\pi -\pi\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{1 -\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{(1 -\sqrt{cos{x}})(1 +\sqrt{cos{x}})}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{1 - cos{x}}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x^2}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x}{(1 +\sqrt{cos{x}})}=0$$
In this answer I will use the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to derive the limit
\begin{align} \lim_{x \to 0} \frac{\sin \left( \pi\sqrt{\cos x} \right)}{x}&=\lim_{x \to 0} \frac{\sin \left(\pi - \pi \sqrt{\cos x} \right)}{x}\\ &= \lim_{x \to 0} \frac{\sin \left(\pi\left(1 - \sqrt{\cos x}\right) \right)}{\pi\left(1 - \sqrt{\cos x}\right) }\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\ &=\lim_{x \to 0}\frac{\pi\left(1 - \sqrt{\cos x}\right) }x\\ &= \lim_{x \to 0} \frac{\pi\left ( 1 - \cos x \right) }{x \left (1 + \sqrt{\cos x}\right)}\\ &= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos x }x\\ &= \frac{\pi}2 \lim_{x \to 0} \frac{1 - \cos^2 x }{x\left ( 1 + \cos x\right) }\\ &= \frac{\pi}4 \lim_{x \to 0} \frac{sin^2 x}{x}\\ &= \frac{\pi}4 \lim_{x \to 0} \sin x \\ &= 0 \end{align}