how to calculate $\lim_{x\rightarrow 0} \frac{(\sin{\pi \sqrt{ \cos (x)}})}{x}$ without L'Hopital?

Question

Calculate the following limit: $$ \lim_{x\rightarrow 0} \frac{(\sin{\pi \sqrt{ \cos (x)}})}{x}$$

I have no idea please help

Answer 1

We rewrite it as a product of factors, all of whose limits are known :

$$ \frac{\sin{(\pi \sqrt{ \cos (x)})}}{x}=-\frac{\sin{(\pi (\sqrt{ \cos (x)}-1))}}{x}$$ $$=-\pi\frac{\sin{(\pi (\sqrt{ \cos (x)}-1))}}{\pi (\sqrt{ \cos (x)}-1)}\frac{\sqrt{ \cos (x)}-1}{x}$$

$$=-\pi\frac{\sin{(\pi (\sqrt{ \cos (x)}-1))}}{\pi (\sqrt{ \cos (x)}-1)}\frac{ \cos (x)-1}{x^2}\frac{1}{\sqrt{ \cos (x)}+1}x$$

Thus I get (if I havent made a mistake) a limit of $0$.

Answer 2

$$ \begin{align} \lim_{x\to0}\frac{\sin\left(\pi\sqrt{\cos(x)}\right)}{x} &=\lim_{x\to0}\frac{\sin\left(\pi\left(1-\sqrt{\cos(x)}\right)\right)}{x}\\ &=\lim_{x\to0}\frac{\sin\left(\pi\left(1-\sqrt{\cos(x)}\right)\right)}{\pi\left(1-\sqrt{\cos(x)}\right)}\lim_{x\to0}\frac{\pi\left(1-\sqrt{\cos(x)}\right)}{x}\\ &=1\cdot\lim_{x\to0}\frac{\pi\left(1-\cos^2(x)\right)}{x\left(1+\sqrt{\cos(x)}\right)(1+\cos(x))}\\ &=\frac\pi4\lim_{x\to0}\frac{\sin^2(x)}{x}\\[3pt] &=\frac\pi4\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\sin(x)\\[4pt] &=\frac\pi4\cdot1\cdot0 \end{align} $$

Answer 3

Let $f(x) = \sin(\pi\sqrt {\cos x}).$ The expression equals

$$\frac{f(x) - f(0)}{x-0}.$$

By definition of the derivative, this $\to f'(0)$ as $x\to 0.$ This equals $\cos(\pi\sqrt {\cos 0})\cdot \pi/(2\sqrt {\cos 0}) \cdot (-\sin 0) = 0.$

Answer 4

Considering $$y=\frac{\sin\left(\pi\sqrt{\cos(x)}\right)}{x}$$ you could get much more than the limit using Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$$ $$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}+O\left(x^5\right)$$ $$\sin\left(\pi\sqrt{\cos(x)}\right)=\frac{\pi }{4}x^2+\frac{\pi }{96}x^4+O\left(x^5\right)$$ $$y=\frac{\sin\left(\pi\sqrt{\cos(x)}\right)}{x}=\frac{\pi }{4}x+\frac{\pi }{96} x^3+O\left(x^4\right)$$