# Evaluating $\lim_{{x \to 0}} \frac{\cos\left(\frac{\pi}{2 \cos x}\right)}{\sin(\sin x^2)}$ without using L'Hôpital's Rule

I'm working on solving the following trigonometric limit without using L'Hôpital's rule and could use some help:

$$\lim_{{x \to 0}} \frac{\cos\left(\frac{\pi}{2 \cos x}\right)}{\sin(\sin x^2)}$$ I've attempted some preliminary simplifications such as using the identity $$\cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right)$$ to reframe parts of the function, but I'm still finding it challenging to proceed further from here. Additionally, I tried to consider the behavior of the functions as $$x$$ approaches zero, but I couldn't draw a concrete conclusion.

Is there a way to approach this limit using algebraic manipulation, trigonometric identities, or some other method that avoids derivatives? Any guidance or hints would be greatly appreciated.

We only need to use two standard limits: $$\lim_{t\to 0}\frac{\sin(t)}{t}=1, \quad \lim_{t\to 0} \dfrac{1-\cos(t)}{t^2/2}=1$$

Using the first limit twice, we find $$\lim_{x\to 0}\frac{\sin(\sin(x^2))}{x^2}=1$$

And using the identity $$\cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right)$$, we get $$\cos\left(\frac{\pi}{2\cos(x)}\right)=\sin\left(\frac{\pi}{2}\left(1-\frac{1}{\cos(x)}\right)\right)=\sin\left(\frac{\pi}{2}\left(\frac{\cos(x)-1}{\cos(x)}\right)\right)$$

Using the first limit, when $$x\to 0$$ we may replace the above expression by $$\frac{\pi}{2}\left(\frac{\cos(x)-1}{\cos(x)}\right),$$

and by the second limit this may be replaced by $$-\frac{\pi}{2}\frac{x^2/2}{\cos(x)}$$

Putting all the pieces together, we conclude $$\lim_{{x \to 0}} \frac{\cos\left(\frac{\pi}{2 \cos x}\right)}{\sin(\sin x^2)}=\lim_{x\to 0}-\frac{\frac{\pi}{2}\frac{x^2/2}{\cos(x)}}{x^2}=\lim_{x\to 0}-\frac{\pi}{4\cos(x)}=-\frac{\pi}{4}$$

@Julio Puerta gave a great answer. I'll try to give a correct answer. $$\lim_{{x \to 0}} \frac{\cos\left(\frac{\pi}{2 \cos x}\right)}{\sin(\sin x^2)}$$ $$=\lim_{x \to 0} \frac{\sin(\frac{π}{2}(1-\frac{1}{\cos x}))}{\sin(\sin x^2)}=\lim_{x\to 0} \frac{\sin(\frac{π}{2}(1-\frac{1}{\cos x}))}{\frac{π}{2}(1-\frac{1}{\cos x})}×\frac{π}{2}(1-\frac{1}{\cos x})×\frac{\sin x^2}{\sin(\sin x^2)}×\frac{x^2}{\sin x^2}×\frac{1}{x^2}$$ Applying $$\lim_{t\to 0} \frac{\sin t}{t}=1$$ thrice, $$=\lim_{x \to 0} \frac{π(\cos x -1)}{2x^2 \cos x}=\frac{π}{2}\lim_{x\to 0} \frac{-2\sin^2\frac{x}{2}}{x^2}$$ We apply the $$\lim_{t\to 0} \frac{\sin t}{t}=1$$ again (funny thing) $$=\frac{π}{2} \lim_{x\to 0} \frac{-1}{2} ×\frac{\sin^2\frac{x}{2}}{(\frac{x}{2})^2}=\frac{-π}{4}$$

Too long for a comment.

Since you already received very good answers for the limit, let me show how we can get more at the price of simple calculations. I admit that you are familiar with Taylor series.

• For the numerator

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$

Use the long division $$\frac \pi{2\cos(x)}=\frac{\pi }{2}+\frac{\pi x^2}{4}+\frac{5 \pi x^4}{48}+O\left(x^6\right)$$ $$\cos\left(\frac \pi{2\cos(x)}\right)=-\frac{\pi x^2}{4}-\frac{5 \pi x^4}{48}+O\left(x^6\right)$$

• For the denominator

$$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^8\right)$$ $$\sin(\sin(x^2))=x^2-\frac{x^6}{3}+O\left(x^8\right)$$

• For the expression

$$\frac{\cos\left(\frac \pi{2\cos(x)}\right)} {\sin(\sin(x^2)) }=\frac{-\frac{\pi x^2}{4}-\frac{5 \pi x^4}{48}+O\left(x^6\right) } {x^2-\frac{x^6}{3}+O\left(x^8\right) }$$

Long division again $$\frac{\cos\left(\frac \pi{2\cos(x)}\right)} {\sin(\sin(x^2)) }=-\frac{\pi }{4}-\frac{5 \pi x^2}{48}+O\left(x^4\right)$$

Put the two functions on the same plot for $$-0.5 \leq x \leq 0.5$$ to see that this is "quite" good.

All of the above has been done by hand.

If you want more accuracy, use Wolfram Alpha.