I've been asked to solve the limit.
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Here's my approach:
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Using the identity, $\cos(x) =\sin(90^{\circ} – x)$
\begin{aligned}\implies \lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
& = \lim_{x\to0}\frac{\sin\left(\frac{\pi}{2} – \frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
\\& = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} – \dfrac{\pi}{2\cos(x)}\right)\cdot\dfrac{\sin\left(\frac{\pi}{2} – \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} – \frac{\pi}{2\cos(x)}\right)}}{\sin(\sin(x^2))}
\\ & = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} – \dfrac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}\cdot \underbrace{\lim_{x\to0}\dfrac{\sin\left(\frac{\pi}{2} – \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} – \frac{\pi}{2\cos(x)}\right)}}_{1}
\\ & = \dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) – 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(\sin(x^2))}{\sin(x^2)}}_1\cdot\sin(x^2)}
\\ & =\dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) – 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(x^2)}{x^2}}_1\cdot x^2}
\\ & = \color{blue}{\boxed{\lim\limits_{x\to0}\dfrac{\pi}{2x^2}\left(\dfrac{\cos(x) – 1}{\cos(x)}\right)}}
\end{aligned}
Now, I'm unable to think of anything to do with this boxed part. Can anyone check my above method and tell me what to do further with this question? Any other shorter method is also most welcomed!
Best Answer
$$ \frac{\cos x - 1}{x^2} = \frac{\cos x - 1}{x^2} \times \frac{\cos x + 1}{\cos x + 1} = \frac{\cos^2 x - 1}{x^2(\cos x + 1)} = \frac{-\sin^2 x}{x^2} \times \frac{1}{\cos x + 1} $$ and the left fraction can be computed since $\sin(x)/x \to 1$...