evaluate the limit $\lim_{x\to 0}\frac{h(x)}{g(x)}$

Question

evaluate this limit $$\lim_{x\to 0}\frac{cos^2(1-cos^2(1-cos^2(1-cos^2(.....1...\cos^2 x)).....)}{sin(\pi(\frac{\sqrt{x+4}-2}{x}))}$$

i solved it by by assuming both the individual limit exists and i used limit quotient rule,am i correct? $$\frac{\lim {h(x)}}{\lim {g(x)}}$$

then i applied l'hopitals rule for $ g(x)$ by assuming it to be continous : $$\sin\pi(\lim_{x\to 0}{\frac{\sqrt{x+4}-2}{x})}$$

we can see it to be $0/0$ form i don't know was this correct?

then i got my answer. now are my steps correct?

and any other way of solving is welcomed.

EDIT

This question i got from comments.Can the numerator have infinitely many iterations of ff? And should its limit still be 1 in this case?

Answer 1

$$ \frac{\sqrt{4+x}-2}x=\frac1{\sqrt{4+x}+2} $$

If the numerator is the limit of the fixed point iteration of $f(u)=\cos^2(1-u)$ then it should be helpful to attempt to show that the unique (positive) fixed point is $u=1$, which should be easy since $1-f(u)=\sin^2(1-u)\le\sin(1)(1-u)$ for all $u\in [0,1]$.

To make it formally correct write the numerator as $f^\infty(x)=\lim_{n\to\infty}f^n(x)$ where $f^n$ is the $n$-fold composition of $f$ with itself. Since $f(x)\in[0,1]$ and $f$ is contractive there, $ f^\infty\equiv 1$ is a constant function.

Answer 2

The numerator is $\cos^2\left(\sin^2\left(\sin^2\left(\dots\sin^2(x)\right)\right)\right)$ which tends to $1$ as $x\to0$.

The denominator is $\sin\left(\pi\frac{\sqrt{4+x}-2}{x}\right)=\sin\left(\pi\frac1{\sqrt{4+x}+2}\right)$ which tends to $\sin\left(\frac\pi4\right)=\frac1{\sqrt2}$ as $x\to0$.

The ratio, therefore, tends to $\sqrt2$.