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. 
 A: $$\frac{\sin(\tan x)}{\sin x} = \frac{\sin(\tan x)}{\tan x} \frac{\tan x}{\sin x} = \frac{\sin(\tan x)}{\tan x}\frac{1}{\cos x}.$$
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 $1\cdot 1 = 1$
A: $$\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)}=\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)/x} \cdot \dfrac{1}{x} = \lim_{x \to 0}\dfrac{\sin(\tan(x))/\tan(x)}{\sin(x)/x} \cdot \dfrac{\tan(x)}{x}\text{.}$$
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
$$\lim_{x \to 0}\tan(x)/x=\sec(0) = 1\text{.}$$
A: 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}
A: $$\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}=\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}\cdot\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}}=\lim_{x\to 0}\frac{\sec x\sin(\tan x)}{\tan x}$$
Can you take it from here?
A: 
Here we present a solution that relies on only (i) elementary inequalities from geometry and (ii) the squeeze theorem.


NOTE:
We first note that $\frac{\sin(\tan (x))}{\sin(x)}$ is an even function of $x$ and hence, if the right-side limit $\lim_{x\to 0^+}\frac{\sin(\tan (x))}{\sin(x)}$ exists, then the limit $\lim_{x\to 0}\frac{\sin(\tan (x))}{\sin(x)}$ exists.  The ensuing analysis focuses, therefore, on establishing the right-side limit.

First, recall from elementary geometry that the sine function satisfies the inequalities
$$x\cos(x)\le \sin(x)\le x \tag 1$$
for $0\le x\le \pi/2$.  From $(1)$ it is easy to see that
$$x\le \tan(x)\le \frac{x}{\cos(x)} \tag 2$$
for $0\le x<\pi/2$.  Using $(1)$ and $(2)$, we can write for $0<x<\arctan(\pi/2)$
$$\cos(\tan(x))\le \frac{\sin(\tan(x))}{\sin(x)}\le \frac{1}{\cos^2(x)} \tag 3$$
Finally, applying the squeeze theorem to $(3)$ yields 
$$\lim_{x\to 0^+} \frac{\sin(\tan(x))}{\sin(x)}=1$$
whereupon exploiting the evenness of $\frac{\sin(\tan(x))}{\sin(x)}$ establishes the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0} \frac{\sin(\tan(x))}{\sin(x)}=1}$$

Tools Used: 
  
  
*
  
*The inequalities in $(1)$ and $(2)$
  
*The Squeeze Theorem
  
*Continuity of the cosine and tangent functions for $|x|<\pi/2$
  

A: You can always try Taylor expansions.
$$\frac{\sin(\tan x)}{\sin x} = \frac{\tan x - \frac{\tan^3 x}{3!} + \mathcal{O}(\tan x)^5}{\sin x} = \frac{x+\frac{x^3}{3!} + \mathcal{O}(x^5)- \frac{\tan^3 x}{3!} + \mathcal{O}(\tan x)^5}{x - \frac{x^3}{3!} + \mathcal{O}(x^5)} = \frac{1+\frac{x^2}{3!} + \mathcal{O}(x^4)}{1 - \frac{x^2}{3!} + \mathcal{O}(x^4)}- \frac{\frac1x\frac{\tan^3 x}{3!} + \frac1x\mathcal{O}(\tan x)^5}{1 - \frac{x^2}{3!} + \mathcal{O}(x^4)} \rightarrow 1 \text{ as } x\rightarrow 0$$
In the last equality, note that the numerator in the latter term can be expanded as a polynomial with a zero constant term. Thus it vanishes in the limit.
