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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.

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5 Answers 5

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$u(0)$ 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$.

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    $\begingroup$ It should be noted that $\tan \, x$ does not vanish for $0<|x|<1$. $\endgroup$ Commented Jun 23, 2019 at 2:04
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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.

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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.

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

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

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