Limit of $\lim \limits_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$ I was trying to solve this problem, but couldn't figure it out. The solution goes like this:

I don't understand the first step. Why is the limit multiplied by $\frac{4x}{5x}$? and $\frac{5}{4}$ ? 
 A: Alternatively, you could use L'Hopital's rule:
$\lim \limits_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$=$\lim \limits_{x \to 0} \frac{\frac{d}{dx}\sin(5x)}{\frac{d}{dx}\sin(4x)}$=$\lim \limits_{x \to 0} \frac{5\cos(5x)}{4\cos(4x)}=\frac{5}{4}$
A: $$\lim \limits_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$$
$$\lim \limits_{x \to 0} \frac{\sin(5x)}{x} . \frac{x}{\sin(4x)}$$
$$∵\lim \limits_{\theta \to 0}\frac{ \sin(a\theta)}{\theta}=a$$
$$=\frac54$$
A: $\lim\limits_{x\to0}\frac{\sin(5x)}{\sin(4x)}=$
$\lim\limits_{x\to0}\frac{\sin(5x)\cdot5x\cdot4x}{\sin(4x)\cdot5x\cdot4x}=$
$\lim\limits_{x\to0}\frac{\sin(5x)\cdot4x\cdot5x}{5x\cdot\sin(4x)\cdot4x}=$
$\lim\limits_{x\to0}\left(\frac{\sin(5x)}{5x}\cdot\frac{4x}{\sin(4x)}\cdot\frac{5x}{4x}\right)=$
$\left(\lim\limits_{x\to0}\frac{\sin(5x)}{5x}\right)\cdot\left(\lim\limits_{x\to0}\frac{4x}{\sin(4x)}\right)\cdot\left(\lim\limits_{x\to0}\frac{5x}{4x}\right)=$
$\left(\lim\limits_{x\to0}\frac{\sin(5x)}{5x}\right)\cdot\left(\lim\limits_{x\to0}\frac{4x}{\sin(4x)}\right)\cdot\left(\lim\limits_{x\to0}\frac{5}{4}\right)=$
$1\cdot1\cdot\frac54=$
$\frac54$
A: They were using the Squeeze theorem but I don't think it is necessary (https://www.khanacademy.org/math/differential-calculus/limits-from-equations-dc/squeeze-theorem-dc/v/proof-lim-sin-x-x)
$$\lim_{x \to 0} \frac{\sin5x}{\sin4x}$$
$$= \frac{\lim_{x \to 0} \sin5x}{\lim_{x \to 0} \sin4x}$$
where $\lim_{x \to 0} \sin4x$ not equal to $0$
but it is equal to zero so we can't do this so we can use L'Hopitals Rule where you differentiate top and bottom. We will be using chain rule here as well because $\frac{d}{dx}\sin5x = \cos5x \times \frac{d}{dx}5x = 5\cos 5x$
$$\lim_{x \to 0} \frac{\sin5x}{\sin4x}$$
$$=^{L'H} \lim_{x \to 0} \frac{5\cos5x}{4\cos4x}$$
Now just substitute zero
$$= \frac{5\cos(5(0))}{4\cos(4(0))}$$
$$=\frac{5(1)}{4(1)}$$
$$=\frac{5}{4}$$
