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. 

$$\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}=\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}\cdot\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}}=\lim_{x\to 0}\frac{\sec x\sin(\tan x)}{\tan x}$$
Can you take it from here?
