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. 

Recall that $\tan x = \frac{\sin x}{\cos x}$ and that $\cos x = \sqrt{1 - \sin^2 x}$. Let $u = \sin x$
\begin{align}
\lim_{x \to 0} \frac{\sin(\frac{\sin x}{\cos x})}{\sin x} &= \lim_{x \to 0} \frac{\sin(\frac{\sin x}{\sqrt{1 - \sin^2 x}})}{\sin x}\\
&= \lim_{u \to 0} \frac{\sin \frac{u}{\sqrt{1-u^2}}}{\frac{u}{\sqrt{1 - u^2}}} \frac{1}{\sqrt{1 - u^2}}\\
&= 1 \cdot \frac{1}{\sqrt{1 - 0}} = 1
\end{align}
