# 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?

$$u(0)$$ 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 $$s\to 0$$.

• It should be noted that $\tan \, x$ does not vanish for $0<|x|<1$. – Kabo Murphy Jun 23 at 2:04

Your reasoning is fine, just define $$u(t)=\begin{cases}\frac{\sin(t)}{t} &, t\neq 0 \\ 1 &, t=0 \end{cases}$$ In this case $$u$$ is continuous at $$0$$ and apply the theorem.

L'Hôpital's rule is applicable. You get $$\lim_{t\to0}\dfrac {\cos (\tan t)\cdot\sec^2t}{\sec^2t}=\lim_{t\to0}\cos (\tan t)=\cos\tan0=\cos0=1$$, where I used continuity twice at the end.

Expressing this in terms of taylor-series, using Pari/GP, it gives

gp> sin(tan(x))/tan(x)
%97 = 1 - 1/6*x^2 - 37/360*x^4 - 787/15120*x^6 - 6013/259200*x^8 - 1108099/119750400*x^10
- 235825223/72648576000*x^12 - 3625913363/3923023104000*x^14 + O(x^16)


which for $$\lim_{x=t\to 0}$$ gives of course $$\small \begin{array} {rlll}\phantom= \lim_{x\to 0} &\sin(\tan(x))/\tan(x) \\ =\lim_{x\to 0} & 1 - 1/6x^2 - 37/360x^4 - 787/15120x^6 - 6013/259200x^8 - 1108099/119750400x^{10} - O(x^{12}) \\ = & 1 \end{array}$$

$$\sin(x)$$ 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 $$\tan(x)/\tan(x)=1$$.