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. 


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 $|x|<\pi/2$
  

