let $a_1=1$ and $a_n=\sin(a_{n-1})$, $n>1$ and $n \in N$ The Calculate the following Limit

Question

Let $a_1=1$ and $a_n=\sin(a_{n-1})$, $n>1$ and $n \in N$ If $$\lim_{n\to\infty} \frac{2^{2{a_n}}-2^{1+a_n} \cdot 3^{a_n}+3^{2a_n}}{\cos(a_n)+1-e^{a_n}-e^{-a_n}}=l\quad$$ Then calculate $l$

How to solve it without Using L'Hopital's Rule.

My Approach:

$$\begin{align} a_2 &=\sin(a_1)\\ \implies a_2 &=\sin(1)\\ \implies a_3 &=\sin(\sin1)\\ \end{align} $$ Similarly $a_n=\sin(\sin(\sin\sin(....(sin1))))$

Which will Approach to Zero for $n \to \ \infty$

$\implies$ $\lim_{n\to\infty} a_n=0$

Later I Replaced $a_n$ with $x$

So question changed to $$\lim_{x\to 0} \frac{2^{x}-2^{1+x} \cdot 3^{x}+3^{2x}}{\cos(x)+1-e^{x}-e^{-x}}=l \implies \lim_{x\to 0} \frac{(2^x-3^x)^2}{1+\cos(x)-e^x-e^{-x}}$$

How to solve it further without using L'Hopital's Rule.

Answer 1

Hint : when $x$ tends to $0$, you have

$$\frac{(2^x-3^x)^2}{1+\cos(x)-e^x-e^{-x}} = \frac{(x \ln(2/3) + o(x))^2}{-\frac{3x^2}{2}+ o(x^2)} = -\frac{2}{3}\ln^2\left( \frac{3}{2}\right) + o(1)$$

so the limit is $$-\frac{2}{3}\ln^2\left( \frac{3}{2}\right)$$