calculus - Stuck on basic limit problem: $lim_{x to 0} frac{sin(tan x)}{sin x}$

Consider $\lim_{x \to 0} \frac{\sin(\tan x)}{\sin x}$. The answer is $1$. This is clear intuitively since $\tan x ≈ x$ for small $x$. How do you show this rigorously? In general, it does not hold that $\lim_{x \to p} \frac{f(g(x))}{f(x)} = 1$ if $g(x) - x \to 0$ as $x \to p$.

No advanced techniques like series or L'Hôpital. This is an exercise from a section of a textbook which only presumes basic limit laws and continuity of composite continuous functions.

This should be a simple problem but I seem to be stuck. I've tried various methods, including $\epsilon-\delta$, but I'm not getting anywhere. The composition, it seems to me, precludes algebraic simplification.

Answer

$$\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)}=\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)/x} \cdot \dfrac{1}{x} = \lim_{x \to 0}\dfrac{\sin(\tan(x))/\tan(x)}{\sin(x)/x} \cdot \dfrac{\tan(x)}{x}\text{.}$$
Now
$$\lim_{x \to 0}\sin(\tan(x))/\tan(x) = \lim_{\tan(x) \to 0}\sin(\tan(x))/\tan(x) = 1\text{,}$$
$$\lim_{x \to 0}\sin(x)/x = 1$$
and you can use this (or any of the other answers if you haven't covered derivatives) to show
$$\lim_{x \to 0}\tan(x)/x=\sec(0) = 1\text{.}$$