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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.

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3 Answers 3

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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 $$\lim_{{x \to 0}} \frac{\cos\left(\frac{\pi}{2 \cos x}\right)}{\sin(\sin x^2)}=\lim_{x\to 0}-\frac{\frac{\pi}{2}\frac{x^2/2}{\cos(x)}}{x^2}=\lim_{x\to 0}-\frac{\pi}{4\cos(x)}=-\frac{\pi}{4}$$

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@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) $$=\frac{π}{2} \lim_{x\to 0} \frac{-1}{2} ×\frac{\sin^2\frac{x}{2}}{(\frac{x}{2})^2}=\frac{-π}{4}$$

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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.

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