Find $\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x}$

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.

• I would try Binomial and Maclaurin series…. Commented Sep 30, 2022 at 23:12
• Try to recognize the definition of a derivative as the limit of a difference quotient here.
– fwd
Commented Sep 30, 2022 at 23:15
• @AdamRubinson sorry that is beyond my actual level and I am not allowed to use any technique that I didn't learn in class Commented Sep 30, 2022 at 23:22
• Have you learned the squeeze/sandwich theorem? Commented Sep 30, 2022 at 23:32
• @fwd cool method. So you get $f'(x)=\frac{-\sin x}{2\sqrt{\cos x}} \pi \cos (\pi\sqrt{\cos x})$ and so the limit is equal to $f'(0)=0$ Commented Oct 1, 2022 at 0:16

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}
• nice, simple and elegent. That's the answer I was looking for, well done! The crux move here was to use $\sin{x} = \sin{(\pi - x)}$, thanks for the answer. Commented Sep 30, 2022 at 23:48
$$sin(\alpha) = sin(\pi - \alpha)$$ $$sin(\pi\sqrt{cos{x}}) = sin(\pi -\pi\sqrt{cos{x}})$$ $$sin(\pi -\pi\sqrt{cos{x}}) \sim_{x\to 0} \pi -\pi\sqrt{cos{x}}$$ $$\lim_{x\to 0} \frac{sin(\pi\sqrt{cos{x}})}{x} = \lim_{x\to 0} \frac{\pi -\pi\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{1 -\sqrt{cos{x}}}{x} = \pi \cdot \lim_{x\to 0} \frac{(1 -\sqrt{cos{x}})(1 +\sqrt{cos{x}})}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{1 - cos{x}}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x^2}{x(1 +\sqrt{cos{x}})} = \pi \cdot \lim_{x\to 0} \frac{0.5x}{(1 +\sqrt{cos{x}})}=0$$
• Does that use the fact that when x approaches 0 $\sin{x} = x$ ? I actually know about this but I only use it in physics to find approximate values for really small angles. What property or theorem lets you use this in maths? Also I didn't understand how you went from $1- \cos x$ to $0.5x^2$. Commented Sep 30, 2022 at 23:54