Limit of composite function: $\lim_{t \to 0} [\frac{\sin(\tan(t))}{\tan(t)}]$ Limit of composite function: $\lim_{t \to 0} [\frac{\sin(\tan(t))}{\tan(t)}]$


This is an exercise from Apostol's chapter on composite functions and continuity, in which the following theorem is stated:


  Let $v$ be continuous on $p$; let $u$ be continuous on $v(p)$, then $u∘v$ is continuous on $p$


So, I'm assuming this type of limits are meant to be evaluated using this theorem.

I define $v(t):=\tan(t)$, and $u(t):=\frac{\sin(t)}{t}$.

It's clear $v$ is continuous on $0$ since $\lim_{t \to 0} \tan(t) = \tan(0) = 0 $
Which means that, to apply the theorem, $\frac{\sin(t)}{t}$ must be continuous on $v(0)$, this is, on $0$. That is not the case, since $u(0)$ isn't even defined. 

However, in this solutions manual they still get to use the theorem, it seems: http://www.stumblingrobot.com/2015/09/02/evaluate-the-given-limit-4/

How should I follow up from here? am I missing something that still allows me to use the theorem? Are they using something else in the solutions manual, and I'm just confused? In case I can't use that theorem here, how should I compute this limit, then?

Thanks in advance.
$u(0)$ is not involved in finding the limit. The required limit is $\lim_{u \to 0} u(s)=1$ because $s =\tan\, t \to 0$ as $t \to 0$ and $u(s) \to 1$ as $ s\to 0$.  

Just apply definition of limit. There is no need to apply any theorem except the one which says  $u(s) \to 1$ as $ s\to 0$.
$\sin(x)$ has the same limit as $x$ when $x$ approaches $0$. This means that $\sin(\tan(x))$ will approach $\tan(x)$ at $x=0$. This means that we are looking at limit as $x$ approaches $0$ of $\tan(x)/\tan(x)=1$.