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

$$\frac{\sin(\tan x)}{\sin x} = \frac{\sin(\tan x)}{\tan x} \frac{\tan x}{\sin x} = \frac{\sin(\tan x)}{\tan x}\frac{1}{\cos x}.$$

As $x\to 0, \tan x \to 0,$ hence the first fraction on the right $\to 1.$ We also know $\cos x \to 1,$ so the second fraction on the right $\to 1.$ The limit is therefore $1\cdot 1 = 1$

$$\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)}=\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)/x} \cdot \dfrac{1}{x} = \lim_{x \to 0}\dfrac{\sin(\tan(x))/\tan(x)}{\sin(x)/x} \cdot \dfrac{\tan(x)}{x}\text{.}$$ 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 $$\lim_{x \to 0}\tan(x)/x=\sec(0) = 1\text{.}$$

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}